11.5/ 11.5 Conic Sections - Cengage

Section 11.5 / Conic Sections 11.5/1

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To graph a parabola

The conic sections are curves that can be constructed from the intersection ofa plane and a right circular cone. The parabola, which was introduced earlier, isone of these curves. Here we will review some of that previous discussion andlook at equations of parabolas that were not discussed before.

Every parabola has an axis of symmetry and a vertex that is on the axis of sym-metry. To understand the axis of symmetry, think of folding the paper along thataxis. The two halves of the curve will match up.

The graph of the equation , is a parabola with the axis ofsymmetry parallel to the y-axis. The parabola opens up when and opensdown when . When the parabola opens up, the vertex is the lowest point onthe parabola. When the parabola opens down, the vertex is the highest point onthe parabola.

The coordinates of the vertex can be found by completing the square.

Find the vertex of the parabola whose equation is y � x2 � 4x � 5.

y � x2 � 4x � 5y � (x2 � 4x) � 5y � (x2 � 4x � 4) � 4 � 5

y � (x � 2)2 � 1

The coefficient of x2 is positive, so the parabolaopens up. The vertex is the lowest point onthe parabola, or the point that has the leasty-coordinate.

Because (x � 2)2 � 0 for all x, the least y-coordi-nate occurs when (x � 2)2 � 0, which occurswhen x � 2. This means the x-coordinate of thevertex is 2.

To find the y-coordinate of the vertex, replace x in y � (x � 2)2 � 1 by 2 andsolve for y.

y � (x � 2)2 � 1� (2 � 2)2 � 1 � 1

The vertex is (2, 1).

a � 0a � 0

y � ax2 � bx � c, a � 0

x

y

Vertex

Axis ofsymmetry

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Point of Interest

Hypatia (c. 340–415) isconsidered the first prominentwoman mathematician. Shelectured in mathematics andphilosophy at the Museum inAlexandria, the mostdistinguished place of learningin the world. One of the topicson which Hypatia lectured wasconic sections. One historianhas claimed that with the death(actually the murder) ofHypatia, “the long and glorioushistory of Greek mathematicswas at an end.”

11.5 Conic Sections

Objective A

HOW TO

y

x– 4 40

– 2

4

– 4

2

– 2 2

Vertex

• Group the terms involving x.• Complete the square on x 2 2 4x. Note that 4 is

added and subtracted. Because 4 2 4 5 0,the equation is not changed.

• Factor the trinomial and combine like terms.

11.5/2 Chapter 11 / Functions and Relations

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By following the procedure of the last example and completing the square on the

equation y � ax2 � bx � c, we find that the x-coordinate of the vertex is .

The y-coordinate of the vertex can then be determined by substituting this valueof x into y � ax2 � bx � c and solving for y.

Because the axis of symmetry is parallel to the y-axis and passes through the ver-

tex, the equation of the axis of symmetry is x 5 .

Find the vertex and axis of symmetry of the parabola whoseequation is y � �3x2 � 6x � 1. Then sketch its graph.

x-coordinate:

The x-coordinate of the vertex is 1.The axis of symmetry is the line x � 1.

To find the y-coordinate of the vertex, replace x by 1 and solve for y.

y � �3x2 � 6x � 1� �3(1)2 � 6(1) � 1 � 4

The vertex is (1, 4).

Because a is negative, the parabola opens down.

Find a few ordered pairs and use symmetry tosketch the graph.

Find the vertex and axis of symmetry of the parabola whoseequation is y � x2 � 2. Then sketch its graph.

x-coordinate:

The x-coordinate of the vertex is 0.The axis of symmetry is the line x � 0.

To find the y-coordinate of the vertex, replace x by 0 and solve for y.

y � x2 � 2� 02 � 2 � �2

The vertex is (0, �2).

Because a is positive, the parabola opens up.


�b2a

� �0

2(1)� 0

�b2a

� �6

2(�3)� 1

�b

2a

�b2a

HOW TO

Point of Interest

The suspension cables forsome bridges, such as theGolden Gate bridge, hang inthe shape of a parabola.Parabolic shapes are also usedfor mirrors in telescopes and incertain antenna designs.

y

x– 4

2

4

4

– 2 0

– 2

– 4

2

y

x– 4 40

4

– 4

– 2

2

2

– 2

• Find the x-coordinate of the vertex andthe axis of symmetry, a 5 23, b 5 6.

• Find the x-coordinate of the vertex andthe axis of symmetry. a 5 1, b 5 0

HOW TO


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.The graph of an equation of the formx � ay2 � by � c, a � 0, is also a parabola. In thiscase, the parabola opens to the right when a is pos-itive and opens to the left when a is negative.

For a parabola of this form, the y-coordinate of the

vertex is . The axis of symmetry is the line

y 5 .

Using the vertical line test, the graph of a parabola of this form is not the graphof a function. The graph of x � ay2 � by � c is a relation.

Find the vertex and axis of symmetry of the parabola whoseequation is x � 2y2 � 8y � 5. Then sketch its graph.

y-coordinate:

The y-coordinate of the vertex is 2.The axis of symmetry is the line y � 2.

To find the x-coordinate of the vertex, replace y by 2 and solve for x.

x � 2y2 � 8y � 5� 2(2)2 � 8(2) � 5 � �3

The vertex is (�3, 2).

Since a is positive, the parabola opens to theright.


Find the vertex and axis of symmetry of the parabola whoseequation is x � �2y2 � 4y � 3. Then sketch its graph.

y-coordinate:

The y-coordinate of the vertex is �1.The axis of symmetry is the line y � �1.

To find the x-coordinate of the vertex, replace y by �1 and solve for x.

x � �2y2 � 4y � 3� �2(�1)2 � 4(�1) � 3 � �1

The vertex is (�1, �1).

Because a is negative, the parabola opens to theleft.


�b2a

� ��4

2(�2)� �1

�b2a

� ��82(2)

� 2

�b

2a

�b

2a

x

y

Vertex

Axis ofsymmetry

y

x– 4

2

40

– 2

4

– 4

2– 2

y

x– 4 2 40– 2

– 2

4

– 4

2

• Find the y-coordinate of the vertex andthe axis of symmetry. a 5 2, b 5 28

• Find the y-coordinate of the vertex andthe axis of symmetry. a 5 22, b 5 24

HOW TO

HOW TO


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Example 1

Find the vertex and axis of symmetry of theparabola whose equation is y � x2 � 4x � 3.Then sketch its graph.

Solution

Axis of symmetry:x � 2

y � 22 � 4(2) � 3� �1

Vertex: (2, �1)

�b2a

� ��42(1)

� 2

You Try It 1


Your solution

Example 2

Find the vertex and axis of symmetry of theparabola whose equation is x � 2y2 � 4y � 1.Then sketch its graph.

Solution

Axis of symmetry:y � 1

x � 2(1)2 � 4(1) � 1� �1

Vertex: (�1, 1)

�b2a

� ��42(2)

� 1

You Try It 2

Find the vertex and axis of symmetry of theparabola whose equation is x � �y2 � 2y � 2.Then sketch its graph.

Your solution

x

y

– 4

4

– 4

– 20 4

2

– 2x

y

– 4 2

4

– 4

– 20 4

2

– 2

x

y

– 4 2

4

– 4

– 20 4

2

– 2x

y

– 4

4

– 4

– 20

2

– 2 2 4

x

y

– 4 42

4

– 4

– 20– 2

2

x

y

2

– 4 4

– 4

0– 2– 2

4

2

Solutions on p. 11.5/S1

Example 3

Find the vertex and axis of symmetry of the parabola whose equation is y � x2 � 1.Then sketch its graph.

Solution


y � 02 � 1� 1

Vertex: (0, 1)

�b2a

� �0

2(1)� 0

You Try It 3


Your solution


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To find the equation of a circle and to graph a circle

A circle is a conic section formed by the intersection of a cone and a planeparallel to the base of the cone.

A circle can be defined as all points (x, y) in the plane that are a fixeddistance from a given point (h, k) called the center. The fixed distance is theradius of the circle.

The Standard Form of the Equation of a Circle

Let r be the radius of a circle and let (h, k) be the coordinates of the center of the circle.Then the equation of the circle is given by

(x � h)2 � (y � k)2 � r 2

Sketch a graph of (x � 1)2 � ( y � 2)2 � 9.

(x � 1)2 � [y � (�2)]2 � 32

Center: (1, �2) Radius: 3

Find the equation of the circle with radius 4 and center (�1, 2).Then sketch its graph.

(x � h)2 � ( y � k)2 � r2

[x � (�1)]2 � ( y � 2)2 � 42

(x � 1)2 � ( y � 2)2 � 16

x

y

4

0

2

– 4

– 2 4– 4– 2

2

x

y

– 4

4

– 20

2

2

– 4

– 2 4

x

y

radiuscenter

r

( )h k,

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• Rewrite the equation in standard form.

• Use the standard form of the equationof a circle.

• Replace r by 4, h by 21, and k by 2.

• Sketch the graph by drawing a circlewith center (21, 2) and radius 4.

HOW TO

HOW TO

As the angle of the planethat intersects the conechanges, different conicsections are formed. Fora parabola, the plane wasparallel to the side of thecone. For a circle, theplane is parallel to thebase of the cone.

T A K E N O T E

Objective B


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Applying the vertical-line test reveals that the graph of a circle is not the graphof a function. The graph of a circle is the graph of a relation.

Example 4


Solution

(x � h)2 � ( y � k)2 � r2

[x � (�2)]2 � ( y � 1)2 � 22

Center: (h, k) � (�2, 1)Radius: r � 2

x

y

2

– 4

4– 4

4

– 20

2

– 2

You Try It 4


Your solution

x

y

– 4

4

2

2– 2– 2

0

– 4

4

Example 5

Find the equation of the circle withradius 5 and center (�1, 3). Then sketchits graph.

Solution

(x � h)2 � ( y � k)2 � r2

[x � (�1)]2 � ( y � 3)2 � 52

(x � 1)2 � ( y � 3)2 � 25

x

y

4

4

6

0

2

– 2– 2

– 4 2

8

You Try It 5

Find the equation of the circle withradius 4 and center (2, �3). Then sketchits graph.

Your solution

y

x– 8 8– 4

8

– 8

4

40

– 4



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.To graph an ellipse with center at the origin

The orbits of the planets around the sun are “oval” shaped. This oval shape canbe described as an ellipse, which is another of the conic sections.

There are two axes of symmetry for an ellipse. The intersection of these twoaxes is the center of the ellipse.

An ellipse with center at the origin isshown at the right. Note that there are twox-intercepts and two y-intercepts.

The Standard Form of the Equation of an Ellipse with Center

at the Origin

The equation of an ellipse with center at the origin is .

The x-intercepts are (a, 0) and (�a, 0). The y-intercepts are (0, b) and (0, �b).

By finding the x- and y-intercepts of an ellipse and using the fact that theellipse is “oval” shaped, we can sketch a graph of the ellipse.

Sketch the graph of the ellipse whose equation is � � 1.

Comparing � � 1 with � � 1, we have a2 � 9 and b2 � 4.

Therefore, a � 3 and b � 2.

The x-intercepts are (3, 0) and (�3, 0).The y-intercepts are (0, 2) and (0, �2).

Use the intercepts to sketch a graph ofthe ellipse.

Using the vertical-line test, we find that the graph of an ellipse is not the graphof a function. The graph of an ellipse is the graph of a relation.

y2

b2

x2

a2

y2

4x2

9

y2

4x2

9

x 2

a 2 �y 2

b 2 � 1

x

y

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x(a, 0)(−a, 0)

(0, −b)

(0, b)

y

x

y

4

0

– 4

4– 4

2

– 2– 2

2

Point of Interest

The word ellipse comes fromthe Greek word ellipsis, whichmeans “deficient.” The methodby which the early Greeksanalyzed the conics caused acertain area in the constructionof the ellipse to be less thananother area (deficient). Theword ellipsis in English, whichmeans “omission,” has thesame Greek root as the wordellipse.

Objective C

HOW TO


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Sketch a graph of the ellipse whose equation is � � 1.

The x-intercepts are (4, 0) and (�4, 0).

The y-intercepts are (0, 4) and (0, �4).

The graph in this example is the graph of a circle. A circle is a special case of

an ellipse. It occurs when a2 � b2 in the equation � � 1.y 2

b2

x2

a2

x

y

0

2

– 2– 2

2

4

– 4

4– 4

y 2

16x2

16

Point of Interest

For a circle, a � b and

thus . Early Greek

astronomers thought thateach planet had a circular orbit.Today we know that the planetshave elliptical orbits. However,in most cases the ellipse is verynearly a circle.

For Earth, . The

most elliptical orbit is Pluto’s,

for which .ab

� 1.0328

ab

� 1.00014

ab

� 1

• a 2 5 16, b 2 5 16

• Use the intercepts and symmetry tosketch the graph of the ellipse.

Example 6

Sketch a graph of the ellipse whose

equation is � � 1.

Solution

x-intercepts:(3, 0) and (�3, 0)

y-intercepts:(0, 4) and (0, �4)

y 2

16x2

9

You Try It 6



Your solution

y 2

25x2

4

Example 7



Solution

x-intercepts:(4, 0) and (�4, 0)

y-intercepts:(0, )and (0, � )( � 3.5)2�3

2�32�3

y 2

12x2

16

You Try It 7



Your solution

y 2

9x2

18

x

y

2 4– 4– 2

0

2

– 4

4

– 2x

y

– 4

4

2

– 20 42– 2

– 4

x

y

2

2– 2

0

4

– 2

– 4

– 4 4x

y

2

– 4

4

– 20

2

– 2 4– 4


HOW TO


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.To graph a hyperbola with center at the origin

A hyperbola is a conic section that is formed by the intersection of a cone and aplane perpendicular to the base of the cone.

The hyperbola has two vertices and an axis of symmetry that passes throughthe vertices. The center of a hyperbola is the point halfway between the twovertices.

The graphs at the right show twopossible graphs of a hyperbola withcenter at the origin.

In the first graph, an axis of sym-metry is the x-axis and the verticesare x-intercepts.

In the second graph, an axis of sym-metry is the y-axis and the verticesare y-intercepts.

Note that in either case, the graph of a hyperbola is not the graph of a function.The graph of a hyperbola is the graph of a relation.

The Standard Form of the Equation of a Hyperbola with

Center at the Origin

The equation of a hyperbola for which an axis of symmetry is the x-axis is

. The vertices are (a, 0) and (�a, 0).

The equation of a hyperbola for which an axis of symmetry is the y-axis is

. The vertices are (0, b) and (0, �b).

To sketch a hyperbola, it is helpful to draw twolines that are “approached” by the hyperbola.These two lines are called asymptotes. As a pointon the hyperbola gets farther from the origin, thehyperbola “gets closer to” the asymptotes.

Because the asymptotes are straight lines, theirequations are linear equations. The equations ofthe asymptotes for a hyperbola with center at

the origin are y � and y � .�ba

xba

x

y 2

b 2 �x 2

a 2 � 1

x 2

a 2 �y 2

b 2 � 1

y

x

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Objective D

y

x

y

x

Point of Interest

The word hyperbola comesfrom the Greek word yperboli,which means “exceeding.” Themethod by which the earlyGreeks analyzed the conicscaused a certain area in theconstruction of the hyperbolato be greater than (to exceed)another area. The wordhyperbole in English, meaning“exaggeration,” has the sameGreek root as the wordhyperbola.

The word asymptote comesfrom the Greek wordasymptotos, which means“not capable of meeting.”

x

y


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Sketch a graph of the hyperbola whose equation is � � 1.

An axis of symmetry is the y-axis.

b2 � 9, a2 � 4

The vertices are (0, 3) and (0, �3).

The asymptotes are y � x and y � x.

y

x8

8

– 8 4

– 8

0– 4

4

– 4

�32

32

x2

4y2

9

• The vertices are (0, b) and (0, 2b).

• The asymptotes are y 5 x andba

y 5 x.�ba

• Sketch the asymptotes. Use symmetry andthe fact that the hyperbola will approachthe asymptotes to sketch its graph.

Point of Interest

Hyperbolas are used in LORAN(LOng RAnge Navigation) as amethod by which a ship’snavigator can determine theposition of the ship, as shownin the figure below. They arealso used as mirrors in sometelescopes to focus incominglight.

T1

T3

T2

Example 8

Sketch a graph of the hyperbola whose


Solution

Axis of symmetry:x-axis

Vertices:(4, 0) and (�4, 0)

Asymptotes:

y � x and y � x�12

12

y 2

4x2

16

You Try It 8



Your solution

y 2

25x2

9

Example 9



Solution

Axis of symmetry:y-axis

Vertices:(0, 4) and (0, �4)

Asymptotes:

y � x and y � x�45

45

x 2

25y 2

16

You Try It 9



Your solution

x2

9y 2

9

y

x8

8

– 8

– 8

4

4

– 4

– 4 0

y

x8

8

– 8

– 8

0

4

– 4

4– 4

y

x8

8

– 8

– 8

4– 4 0

4

– 4

y

x8

8

– 8

– 8

0 4– 4

– 4


HOW TO


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.11.5 Exercises

Objective A To graph a parabola

State (a) whether the axis of symmetry is a vertical or a horizontal line and(b) in what direction the parabola opens.

01. 02. 03.

04. 05. 06.

Find the vertex and axis of symmetry of the parabola given by the equation.Then sketch its graph.

07. x � y2 � 3y � 4 08. y � x2 � 2 09. y � x2 � 2

10. x � 11. x � 12. x �

13. x � 14. y � 15. y �

x

y

2

– 4

4

– 4

420– 2

– 2

4

y

x– 8 80

– 4

8

4– 4

– 8

x

y

2

4

– 4

4– 2 0– 2

2– 4

12

x2 � x � 3�12

x2 � 2x � 6�12

y2 � 2y � 3

x

y

2

– 4

4

– 4

4– 2 0– 2

2x

y

2

– 4

4

– 4

4– 2 0– 2

2x

y

2

– 4

4

– 4

4– 2 0– 2

2

12

y2 � y � 1�14

y2 � 1�12

y2 � 4

x

y

– 4

4

– 4

42– 2 0– 2

2

x

y

2

– 4

4

– 4

42– 2 0– 2

y

x– 8 80

– 4

8

4

4

– 8

– 4

y �14

x2 � 6x � 1x � �12

y2 � 4y � 7x � �3y2 � y � 9

x � y2 � 2y � 8y � �x2 � 5x � 2y � 3x2 � 4x � 7


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22. Find the equation of the circle withradius 2 and center (2, �1). Then sketch itsgraph.

24. Find the equation of the circle with radiusand center (�1, 1). Then sketch its

graph.

23. Find the equation of the circle with radius 3and center (�1, �2). Then sketch its graph.

25. Find the equation of the circle with ra-dius and center (�2, 1). Then sketchits graph.

x

y

4

2 4– 2

– 4

– 4

2

– 2 0

�5

x

y

4

2

– 20

– 4

4– 2– 4 2

x

y

– 4 – 2

4

2

2– 2

0

– 4

4

�5

x

y

– 4

4

– 20

2

– 4

– 2 2 4

Objective B To find the equation of a circle and to graph a circle

Sketch a graph of the circle given by the equation.

16. (x � 2)2 � ( y � 2)2 � 9 17. (x � 2)2 � ( y � 3)2 � 16 18. (x � 3)2 � ( y � 1)2 � 25

19. (x � 2)2 � ( y � 3)2 � 4 20. (x � 2)2 � ( y � 2)2 � 4 21. (x � 1)2 � ( y � 2)2 � 25y

x8

8

– 8

4

0– 8

– 4

4– 4x

y

– 4

4

2

– 4

42– 2

– 2 0x

y

– 4

4

– 20

2

– 4

– 2 42

y

x4 8

8

– 8

4

0– 4– 8

– 4

y

x– 8 4 8

8

– 8

4

– 4

0– 4x

y

– 4

4

2

2– 2 4– 2

0

– 4


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.Objective C To graph an ellipse with center at the origin

Sketch a graph of the ellipse given by the equation.

26. 27. 28.

29. 30. 31.

32. 33. 34.

Objective D To graph a hyperbola with center at the origin

Sketch a graph of the hyperbola given by the equation.

35. 36. 37.

y

x8

8

– 8 4

– 8

0– 4

– 4

4

y

x8

8

4

– 8

– 4

– 8

0 4– 4

y

x8

8

– 8

4

– 8

– 4

– 4 40

y2

16�

x2

9� 1

x2

25�

y2

4� 1

x2

9�

y2

16� 1

y

x– 8 4 8

8

– 8

0– 4

4

– 4

y

x– 8 8

8

– 8

4

– 4

0 4– 4

y

x8

8

– 8

– 8

0

4

– 4

4– 4

x2

4�

y2

25� 1

x2

25�

y2

36� 1

x2

16�

y2

49� 1

y

x8– 8 0

4

– 4

4– 4

8

– 8

y

x8

8

– 8

– 8

0 4– 4

4

– 4

y

x8

8

– 8

– 8

0

4

– 4

4– 4

x2

49�

y2

64� 1

x2

36�

y2

16� 1

x2

16�

y2

9� 1

y

x8

8

– 8

– 8

0

4

– 4

4– 4

y

x8

8

– 8

– 8

0 4– 4

4

– 4

y

x8

8

– 8

– 8

4– 4 0

4

– 4

x2

25�

y2

9� 1

x2

25�

y2

16� 1

x2

4�

y2

9� 1


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38. 39. 40.

41. 42. 43.

44. 45. 46.

APPLYING THE CONCEPTS

Write the equation in standard form. Identify the graph, and then graph theequation.

47. 48. 49.

50. 51. 52.y

x8

8

– 8

– 8 40

4

– 4

– 4

y

x8

8

– 8 4

– 8

– 4

4

0

– 4

y

x– 8 4 8

8

– 4

4

– 8

– 4

0

4y2 � x2 � 369y2 � 16x2 � 14425y2 � 4x2 � �100

y

x8

8

4

– 8

– 4

– 8

4– 4 0

y

x8

8

4

– 8

– 4

– 8

0 4– 4

y

x4

4

– 4

– 4 20

2

– 2

– 2

9x2 � 25y2 � 22516x2 � 25y2 � 4004x2 � 9y2 � 36

y

x– 8 4 8

8

– 4

4

– 8

– 4

0

y

x– 8 8

8

– 8

4

– 4

4– 4 0

y

x8

8

4

– 8

– 8

– 4 4

– 4

0

y2

25�

x2

4� 1

y2

9�

x2

36� 1

x2

36�

y2

9� 1

y

x8

8

– 8

– 8

– 4 40

4

– 4

y

x8

8

4

– 8

– 8

– 4

0– 4 4

y

x8

8

4

– 8

– 8

– 4 4

– 4

0

x2

4�

y2

25� 1

y2

4�

x2

16� 1

y2

25�

x2

9� 1

y

x8

8

4

– 8

– 8

– 4 4

– 4

0

y

x8

8

4

– 8

– 8

– 4

0– 4 4

y

x8

8

– 8

0– 8 4– 4

– 4

4

x2

9�

y2

49� 1

x2

16�

y2

4� 1

y2

16�

x2

25� 1

Solutions to You Try It 11.5/S1

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.SECTION 11.5

You Try It 1

y � x2 � 2x � 1

Axis of symmetry:x � �1

y � (�1)2 � 2(�1) � 1� 0

Vertex: (�1, 0)

You Try It 2

x � �y2 � 2y � 2

Axis of symmetry: y � �1

x � �(�1)2 � 2(�1) � 2� 3

Vertex: (3, �1)

You Try It 3

y � x2 � 2x � 1


y � 12 � 2(1) � 1� �2

Vertex: (1, �2)

You Try It 4

(x � h)2 � (y � k)2 � r2

(x � 2)2 � [y � (�3)]2 � 32

Center: (h, k) � (2, �3)Radius: r � 3

You Try It 5

(x � h)2 � (y � k)2 � r2

(x � 2)2 � [ y � (�3)]2 � 42

(x � 2)2 � ( y � 3)2 � 16

You Try It 6

x-intercepts: (2, 0) and (�2, 0)

y-intercepts: (0, 5) and (0, �5)

You Try It 7

x-interceptsand

y-intercepts:(0, 3) and (0, �3)

You Try It 8

Axis of symmetry: x-axis

Vertices: (3, 0) and (�3, 0)

Asymptotes:

y � x and y � x

You Try It 9

Axis of symmetry: y-axis

Vertices: (0, 3) and (0, �3)

Asymptotes: y � x and y � �x

�53

53

�3�2 ≈ 414�

(�3�2, 0)(3�2, 0)

�b

2a� �

�22(1)

� 1

�b

2a� �

�22(�1)

� �1

�b

2a� �

22(1)

� �1

x

y

– 4 2

4

– 4

– 20 4

2

– 2

x

y

– 4 2

4

– 4

– 20 4

2

– 2

x

y

2

– 4 4

– 4

0– 2– 2

4

2

x

y

– 4

4

2

2– 2– 2

0

– 4

4

y

x– 8 8– 4

8

– 8

4

40

– 4

x

y

– 4

4

2

– 20 42– 2

– 4

x

y

2

2– 2

0

4

– 2

– 4

– 4 4

y

x8

8

– 8

– 8

4– 4 0

4

– 4

y

x8

8

– 8

– 8

4

4

– 4

– 4 0

Answers to Selected Exercises 11.5/A1

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.SECTION 11.5

1a. A vertical line b. Opens up 3a. A horizontal line b. Opens right 5a. A horizontal line b. Opens left

7. 9. 11.

Vertex: Vertex: (0, 2) Vertex: (�1, 0)

Axis of symmetry: Axis of symmetry: x � 0 Axis of symmetry: y � 0

13. 15. 17. 19.

Vertex: (�1, 2)Vertex:

Axis of symmetry: y � 2 Axis of symmetry: x � �1

21. 23. 25. 27.

29. 31. 33. 35.

37. 39. 41. 43.

45. 47. 49. 51.

Ellipse Hyperbola Hyperbola

y

x8

8

– 8 4

– 8

– 4

4

0

– 4

y

x8

8

4

– 8

– 4

– 8

4– 4 0

y

x4

4

– 4

– 4 20

2

– 2

– 2

y2

16�

x2

9� 1

x2

25�

y2

9� 1

x2

9�

y2

4� 1y

x– 8 8

8

– 8

4

– 4

4– 4 0

y

x8

8

– 8

– 8

– 4 40

4

– 4

y

x8

8

4

– 8

– 8

– 4 4

– 4

0

y

x8

8

4

– 8

– 8

– 4

0– 4 4

y

x8

8

– 8 4

– 8

0– 4

– 4

4

y

x8

8

– 8

4

– 8

– 4

– 4 40

y

x8

8

– 8

– 8

0 4– 4

4

– 4

y

x8

8

– 8

– 8

0 4– 4

4

– 4

y

x8

8

– 8

– 8

0 4– 4

4

– 4

x

y

– 4

4

2

– 4

42– 2

– 2 0x

y

– 4

4

2

– 4

42– 2

– 2 0

y

x8

8

– 8

– 8

0 4– 4

4

– 4

(x � 2)2 � (y � 1)2 � 5(x � 1)2 � (y � 2)2 � 9y

x– 8 80– 4

– 4

8

– 8

4

4

��1, �72�

x

y

– 4

4

2

– 4

42– 2

– 2 0

y

x– 8 80– 4

– 4

8

– 8

4

4

x

y

2

– 4

4

– 4

420– 2

– 2x

y

2

4

– 4

4– 2 0– 2

2– 4

y �32

�� 254

, 32�

x

y

2

– 4

4

– 4

4– 2 0– 2

2x

y

– 4

4

– 4

42– 2 0– 2

2

y

x– 8 80– 4

– 4

8

– 8

4

4

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11.5/ 11.5 Conic Sections - Cengage