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State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 1
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 2
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 3
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 4
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 5
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 6
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 7
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 8
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 9
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 10
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 11
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 12
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 13
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 14
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 15
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 16
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
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Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 18
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 19
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 20
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 21
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 22
Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false, replace the underlined term to make a true sentence.
1. The set of all points in a plane that are equidistant from a given point in the plane, called the focus, forms a circle.
SOLUTION: false, center
2. A(n) ellipse is the set of all points in a plane such thatthe sum of the distances from the two fixed points is constant.
SOLUTION: true
3. The endpoints of the major axis of an ellipse are the foci of the ellipse.
SOLUTION: false, vertices
4. The radius is the distance from the center of a circle to any point on the circle.
SOLUTION: true
5. The line segment with endpoints on a parabola, through the focus of the parabola, and perpendicular to the axis of symmetry is called the latus rectum.
SOLUTION: true
6. Every hyperbola has two axes of symmetry, the transverse axis and the major axis.
SOLUTION: false, conjugate axis
7. A directrix is the set of all points in a plane that are equidistant from a given point in the plane, called the center
SOLUTION: false, circle
8. A hyperbola is the set of all points in a plane such that the absolute value of the sum of the distances from any point on the hyperbola to two given points isconstant.
SOLUTION: false; difference
9. A parabola can be defined as the set of all points in aplane that are the same distance from the focus and a given line called the directrix.
SOLUTION: true
10. The major axis is the longer of the two axes of symmetry of an ellipse.
SOLUTION: true
Find the midpoint of the line segment with endpoints at the given coordinates.
11. (–8, 6), (3, 4)
SOLUTION:
Substitute –8, 3, 6 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
12. (–6, 0), (–1, 4)
SOLUTION:
Substitute –6, –1, 0 and 4 for x1, x2, y1 and y2
respectively in the midpoint formula.
13.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the midpoint formula.
14. (15, 20), (18, 21)
SOLUTION:
Substitute 15, 18, 20 and 21 for x1, x2, y1 and y2
respectively in the midpoint formula.
Find the distance between each pair of points with the given coordinates.
15. (10, –3), (1, –5)
SOLUTION:
Substitute 10, 1 –3 and –5 for x1, x2, y1 and y2
respectively in the distance formula.
16. (0, 6), (–9, 7)
SOLUTION:
Substitute 0, –9, 6 and 7 for x1, x2, y1 and y2
respectively in the distance formula.
17.
SOLUTION:
Substitute for x1, x2, y1 and y2
respectively in the distance formula.
18. (5, –3), (7, –1)
SOLUTION:
Substitute 5, 7, –3 and –1 for x1, x2, y1 and y2
respectively in the distance formula.
19. HIKING Marc wants to hike from his camp to a waterfall. The waterfall is 5 miles south and 8 miles east of his campsite. a. How far away is the waterfall? b. Marc wants to stop for lunch halfway to the waterfall. Where should he stop?
SOLUTION: a. The coordinates are (0, –5) and (8, 0). The distance between any two points is
.
Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the distance formula.
b. Substitute 0, 8, –5 and 0 for x1, x2, y1 and y2
respectively in the midpoint formula to find the halfway to the waterfall.
Graph each equation.
20.
SOLUTION:
vertex: (–4, –58)
axis of symmetry: x = –4.
focus:
directrix: y =
length of latus rectum: 1 unit opens up
21.
SOLUTION:
vertex:
axis of symmetry: x = –4.
focus:
directrix: x =
length of latus rectum: 3 unit opens up
22.
SOLUTION:
vertex:
axis of symmetry: y = 4.
focus:
directrix: x =
length of latus rectum: 2 units opens to the right
23.
SOLUTION:
vertex: (–24, 7)
axis of symmetry: y = 7.
focus: (–23.75, 7)
directrix: x = –24.25 length of latus rectum: 1 unit opens to the right
Write each equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.
24.
SOLUTION:
vertex: (0, 0)
axis of symmetry: x = 0opens up
25.
SOLUTION:
vertex: (2, –7)
axis of symmetry: x = 2 opens up
26.
SOLUTION:
vertex: (–5, –3)
axis of symmetry: y = –3opens to the right
27.
SOLUTION:
vertex: (–29, –7)
axis of symmetry: y = –7 opens to the right
28. SPORTS When a football is kicked, the path it travels is shaped like a parabola. Suppose a football is kicked from ground level, reaches a maximum height of 50 feet, and lands 200 feet away. Assumingthe football was kicked at the origin, write an equation of the parabola that models the flight of the football.
SOLUTION: The coordinates of the ball when the ball touches the ground are (0, 0) and (200, 0). The vertex of the path is (100, 50). The standard form of equation of a parabola is
.
Substitute 0, 0 , 100 and 50 for y , x, h and k in the standard form then solve for a.
The equation of the path is
.
Write an equation for the circle that satisfies each set of conditions.
29. center (–1, 6), radius 3 units
SOLUTION: Substitute –1, 6 and 3 for h, k and d in the standard form of circle equation.
30. endpoints of a diameter (2, 5) and (0, 0)
SOLUTION:
The center is .
The radius of the circle is
Substitute 1, 2.5 and for h, k and d in the
standard form of circle equation.
31. endpoints of a diameter (4, –2) and (–2, –6)
SOLUTION:
The radius of the circle is units.
Substitute 1, –4 and 5 for h, k and d in the standard form of circle equation.
Find the center and radius of each circle. Then graph the circle.
32.
SOLUTION:
The center of the circle is (–5, 0).The radius of the circle is 3.
33.
SOLUTION:
The center of the circle is (3, –1). The radius of the circle is 5.
34.
SOLUTION:
The center of the circle is (–2, 8). The radius of the circle is 1.
35.
SOLUTION:
The center of the circle is (–2, 1). The radius of the circle is 4.
36. SOUND A loudspeaker in a school is located at the point (65, 40). The speaker can be heard in a circle with a radius of 100 feet. Write an equation to represent the possible boundary of the loudspeaker sound.
SOLUTION: The center of the circle is (65, 40) and the radius is 100. Substitute 65, 40 and 100 for h, k and r in the standard form of the circle equation.
Find the coordinates of the center and foci and the lengths of the major and minor axes for the ellipse with the given equation. Then graph the ellipse.
37.
SOLUTION:
Here, (h, k) = (0, 0), a = 6 and b = 3.
center:
foci:
length of the major axis: 12 units length of the minor axis: 6 units
38.
SOLUTION:
The values of h, k , a and b are 0, 0, and .
center:
foci:
length of the major axis: units
length of the minor axis: units
39.
SOLUTION:
Center: (h, k) = (0, 4) Here a = 6 and b = 2.
center:
foci:
length of the major axis: 12 unitslength of the minor axis: 4 units
40.
SOLUTION:
Here .
center:
foci:
length of the major axis: 6 units
length of the minor axis: units
41.
SOLUTION:
Here a = 5 and b = 4. center:
foci:
length of the major axis: 10 units length of the minor axis: 8 units
42.
SOLUTION:
Here a = 3 and b = 2.
center:
foci:
length of the major axis: 6 units length of the minor axis: 4 units
43.
SOLUTION:
Here a = 5 and b = 3.
center:
foci:
length of the major axis: 10 units length of the minor axis: 6 units
44.
SOLUTION:
The values of h, k , a and b are 2, 2, .
center:
foci:
length of the major axis: units
length of the minor axis: units
45. LANDSCAPING The Martins have a garden in their front yard that is shaped like an ellipse. The major axis is 16 feet and the minor axis is 10 feet. Write an equation to model the garden. Assume the origin is at the center of the garden.
SOLUTION: The length of the major axis is 16 ft. Therefore, a = 8. The length of the minor axis is 10 ft. Therefore, b = 5. Substitute 8 and 5 for a and b respectively in the standard form of the ellipse equation.
Graph each hyperbola. Identify the vertices, foci, and asymptotes.
46.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (0, ±3).
The foci are .
The equations of the asymptotes are
.
47.
SOLUTION:
The values of h, k , a and b are 3, –2, 1 and 2.
The vertices of the hyperbola are (2, –2) and (4, –2).
The foci are .
The equations of the asymptotes are
.
48.
SOLUTION:
The values of h, k , a and b are 4, –1, 4 and 3.
The vertices of the hyperbola are (4, –5) and (4, 3). The foci are (4, –6) and (4, 4). The equations of the asymptotes are
.
49.
SOLUTION:
The values of h, k , a and b are 0, 0, 3 and 2.
The vertices of the hyperbola are (±3, 0).
The foci are .
The equations of the asymptotes are .
50.
SOLUTION:
The values of h, k , a and b are –2, –1, and 1.
The vertices of the hyperbola are .
The foci are .
The equations of the asymptotes are
.
51. MIRRORS A hyperbolic mirror is shaped like one branch of a hyperbola. It reflects light rays directed at one focus toward the other focus. Suppose a hyperbolic mirror is modeled by the upper branch of
the hyperbola . A light source is located
at (–10, 0). Where should the light hit the mirror so that the light will be reflected to (0, –5)?
SOLUTION:
Foci:
Find the equation of the line joining the points (0, 5) and (–10, 0).
Find the intersection of the line
.
Since the mirror is shaped like one branch of hyperbola and the orientation of the hyperbola is
vertical, the value of x is .
Substitute for x in the equation
.
The light should hit the mirror at
.
Write each equation in standard form. State whether the graph of the equation is a parabola,circle, ellipse, or hyperbola. Then graph.
52.
SOLUTION:
Given equation is a parabola. Graph the function.
53.
SOLUTION:
Given equation is an ellipse. Graph the function.
54.
SOLUTION:
Given equation is a circle. Graph the function.
55.
SOLUTION:
Given equation is an ellipse. Graph the function.
Without writing the equation in standard form, state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.
56.
SOLUTION:
A = 7, B = 0, and C = 9 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse.
57.
SOLUTION:
A = –13, B = 0, and C = 5 The discriminant is
.
Because the discriminant is greater than 0, the conic is a hyperbola.
58.
SOLUTION:
A = 1, B = 0, and C = 0
The discriminant is .
Because the discriminant is equal to 0, the conic is a parabola.
59.
SOLUTION:
A = 1, B = 0, and C = 1
The discriminant is .
Because the discriminant is less than 0, B = 0 and A = C, the conic is a circle.
60. LIGHT Suppose the edge of a shadow can be represented by the equation
.
a. What is the shape of the shadow? b. Graph the equation.
SOLUTION:
a.
A = 16, B = 0, and C = 25 The discriminant is
.
Because the discriminant is less than 0 and A ≠ C, the conic is an ellipse. b. Graph the equation.
Solve each system of equations.
61.
SOLUTION: Solve the linear equation for y .
Substitute –x for y in the quadratic equation and solve for x.
Substitute the values of x in the linear equation and find the value of y .
The solutions are (2, –2) and (–2, 2).
62.
SOLUTION: Solve the linear equation for x.
Substitute 2 + 2y for x in the quadratic equation and solve for y .
Substitute the values of y in the linear equation and find the value of x.
The solution is (–2, –2).
63.
SOLUTION: Solve the linear equation for y .
Substitute for y in the quadratic equation andsolve for x.
Substitute the values of x in the linear equation and find the value of y .
The solution is (4, 0).
64.
SOLUTION:
Multiply the first by 4.
Substitute the values of x in the first equation and find the values of y .
The solutions are (–2, –1), (–2, 1), (2, –1) and (2, 1).
65.
SOLUTION:
Substitute the values of x in the first equation and find the values of y .
The solutions are (1, ±5), (–1, ±5).
66.
SOLUTION: Substitute x for y in the first equation and solve for y .
Substitute the values of x in the linear equation and find the values of y .
The solutions are .
67. PHYSICAL SCIENCE Two balls are launched intothe air at the same time. The heights they are launched from are different. The height y in feet of
one is represented by where t is the time in seconds. The height of the other ball is
represented by .
a. After how many seconds are the balls at the sameheight? b. What is this height?
SOLUTION: a. Equate the both the equations.
b. Substitute 1.5 for t in any one of the equation then evaluate.
The height is 109 ft.
68. ARCHITECTURE An architect is building the front entrance of a building in the shape of a parabola
with the equation . While the
entrance is being built, the construction team puts in two support beams with equations
y = –x + 10 and y = x – 10. Where do the support beams meet the parabola?
SOLUTION: Substitute –x + 10 for y in the quadratic equation andsolve for x.
Substitute the value of x in the equation y = –x + 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point is (0, 10). Substitute x – 10 for y in the quadratic equation and solve for x.
Substitute the value of x in the equation y = x – 10 and find the value of y .
The value of x and y should be positive. Therefore, the intersection point be (20, 10). Therefore, the support beams meet the parabola at (0, 10) and (20, 10).
Solve each system of inequalities by graphing.
69.
SOLUTION:
70.
SOLUTION:
71.
SOLUTION:
72.
SOLUTION:
73.
SOLUTION:
74.
SOLUTION:
eSolutions Manual - Powered by Cognero Page 23
Study Guide and Review - Chapter 9
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