Chapter 10

Conic SectionsConic Sections

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Circle Ellipse Hyperbola

In this chapter, we will study the conic sections. When a right circular cone is intersected bya plane, the result is a conic section. The conic sections are parabolas, circles, ellipses, and hyperbolas. The following figures show how each conic section is obtained from the intersection of a cone and a plane.

Parabola

In Chapter 8, we learned how to graph parabolas. The graph of a quadratic function, f(x)=ax2+bx +c, is a parabola that opens vertically. Another form this function may take is f(x)=a(x – h)2 +k. The graph of a quadratic equation of the form x=ay2 + by + c , or x = a(y – k )2 + h, is a parabola that opens horizontally. The next conic section we will discuss is the circle.

Define a Conic Section

The midpoint of a diameter of a circle is the center of the circle.

Graph a Circle Given in the Form (x – h )2 + (y – k )2 = r2

A circle is defined as the set of all points in a plane equidistant (the same distance) from a fixed point. The fixed point is the center of the circle. The distance from the center to a point on the circle is the radius of the circle.

Let the center of a circle have coordinates (h, k) and let (x, y)represent any point on the circle. Let r represent the distancebetween these two points, r is the radius of the circle. We will use the distance formula to find the distance between the center, (h, k), and the point (x, y) on the circle.

We will use the distance formula to find the distancebetween the center, (h, k), and the point (x, y) on the circle.

212

212 )( yyxxd

Substitute (x, y) for (x2, y2), (h, k) for (x1, y1), and r for d.

22)( kyhxr

222 )( kyhxr

Distance Formula

Square both sides.

This is the standard form for the equation of a circle.

Example 2

Graph (x + 1)2 + (y + 2)2 = 4.SolutionStandard form is (x – h)2 + (y – k)2 = r2.

Our equation is (x + 1)2 + (y + 2)2 = 4.h = -1 k = -2 r = = 24

The center is (-1, -2). The radius is 2. To graphthe circle, first plot the center (-1,-2) use the radius to locate four points on the circle. From the center, move 2 units up, down, left, and right. Draw a circle through the four points.

Example 3

Graph x2 + y2 = 1.SolutionStandard form is (x – h)2 + (y – k)2 = r2.

Our equation is x2 + y2 = 1.h = 0 k = 0 r = = 11

The center is (0,0). The radius is 1. To graphthe circle, first plot the center (0,0) use the radius to locate four points on the circle. From the center, move 1 unit up, down, left, and right. Draw a circle through the four points.

The circle x2 + y2 = 1 is used often in other areasof mathematics such as trigonometry. x2 + y2 = 1 is called the unit circle.

If we are told the center and radius of a circle, we can write its equation.

Example 4

Find an equation of the circle with center (3, -2) and radius .6Solution

The x-coordinate of the center is h. What is h?

The y-coordinate of the center is k. What is k?

What is r?

h = 3

k = -2

r = 6

Substitute these values into (x – h)2 + (y – k)2 = r2

[x – (3)]2 + (y – -2)2 = ( )26

(x – 3)2 + (y + 2)2 = 6

Substitute 3 for x, -2 for k, and for r.6

Equation

Graph a Circle of the Form Ax2 + Ay2 + Cx + Dy + E = 0

The equation of a circle can take another form—general form.

To graph a circle given in this form, we complete the square on x and on y to put it into standard form. After we learn all of the conic sections, it is very important that we understand how to identify each one. To do this, we will usually look at the coefficients of the square terms.

Example 5

Graph

.0134822 yxyx

The coefficients of x2and y2 are each 1. Therefore, this is the equation of a circle. Our goal is to write the given equation in standard form, (x – h)2 + (y – k)2 = r2, so that we can identify its center and radius. To do this, we will group x2 and 8x together, group y2 and 4y together, then complete the square on each group of terms.

Solution

0134822 yxyx 1348 22 yyxx

Group x2 and 8x together.

Group y2 and 4y together.Move the constant to the otherside.

Complete the square for each group of terms.

(x2 + 8x + 16) + (y2 +4y +4) = -13 +16 +4

(x+4)2 + (y+2)2 = 7

The center of the circle is (– 4, – 2 ).The radius is 7

Since 16 and 4 are added on the left, they mustalso be added on the right. Factor.

The center of the circle is (– 4, – 2 ).The radius is 7

(– 4, – 2 )

First plot the center.

(– 4, – 2 )

Second use the radius to plotmore points on the circle. From the center go unitsTo the left, right, up, and down.

65.27

y

x

The next conic section we will study is the ellipse. An ellipse is the set of all points in a plane such that the sum of the distances from a point on the ellipse to two fixed points is constant. Each fixed point is called a focus (plural: foci). The point halfway between the foci is the center of the ellipse.

The EllipseGraph an Ellipse

The orbits of planets around the sun as well as satellites around the earth are elliptical. Statuary Hall in the U.S. Capitol building is an ellipse. If a person stands at one focus of this ellipse and whispers, a person standing across the room on the other focus can clearly hear what was said. Properties of the ellipse are used in medicine as well. One procedure for treating kidney stones involves immersingthe patient in an elliptical tub of water. The kidney stone is at one focus, while at the other focus, high energy shock waves are produced, which destroy the kidney stone.

Graph

Example 1

2 2

4 11

25 9

x y

Solution

Standard form is

12

2

2

2

b

ky

a

hx

Our equation is 2 2

4 11

25 9

x y

What is h? What is k?

What is a? What is b?

h = -4

a= 525 b = 39

Since a = 5 and a2 is under the squared quantity containing the x, move 5 units each way in the x – direction from the center. These are two points on the ellipse.

k = -1

(– 4, – 1 )

y

x

The center is (-4, -1).

To graph the ellipse, first plot the center (-4,-1).

Since b = 3 and b2 is under the squared quantity containing y, move 3 units each way in the y – direction from the center. These are twomore points on the ellipse.

5 5

3

3

Graph

Example 2

1164

22

yx

Solution

Since a = 2 and a2 is under the squared quantity containing the x, move 2 units each way in the x – direction from the center. These are two points on the ellipse.

Standard form is

12

2

2

2

b

ky

a

hx

Our equation is 1164

22

yx

What is h? What is k?

What is a? What is b?

h = 0

a= 416 b = 24

k =0

The center is (0, 0).

To graph the ellipse, first plot the center (0,0).

Since b = 4and b2 is under the squared quantity containing y, move 4 units each way in the y – direction from the center. These are twomore points on the ellipse.

(0, 0 )

y

x

(0, – 4 )

(0, 4)

(2, 0)(– 2, 0)

In Example 2, note that the origin, (0,0), is the center of the ellipse. Notice alsothat a = 2 and the x-intercepts are (2,0) and (-2,0); b = 4 and the y-intercepts are(0,4) and (0, -4). We can generalize these relationships as follows.

Looking at Examples 1 and 2, we can make another interesting observation.

Example 1

2 2

4 11

25 9

x y

a2 = 25 b2 = 9

a2 > b2

1164

22

yx

a2 = 4 b2 = 16

b2 > a2

The number under ( x+4)2 is greaterthan the number under (y+ 1)2. The ellipse is longer in the x-direction.

The number under y2 is greaterthan the number under x2. The ellipse is longer in the y-direction.

This relationship between a2 and b2 will always produce the same result.

The equation of an ellipse can take other forms.

Example 2

Graph

Example 3

144169 22 yxSolutionHow can we tell whether this is a circle or an ellipse? We look at the coefficients of x2 and y2. Both of the coefficients are positive, and they are different. This is an ellipse. (If this were a circle, the coefficients would be the same.)Since the standard form for the equation of an ellipse has a 1 on one side of the sign, divide both sides of by 144to obtain a 1 on the right.

2 2

9 16 144x y

144169 22 yx

144

144

144

16

144

9 22

yx

1916

22

yx

Divide both sides by 144.

Simplify.

The center is (0, 0).

What is a? What is b?

a= 416 b = 39

(0, 0 )

y

x

(0, – 3 )

(0,3)

4, 0)(– 4, 0)

The HyperbolaGraph a Hyperbola

The last of the conic sections is the hyperbola. A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two fixed points is constant. Each fixed point is called a focus. The point halfway between the foci is the center of the hyperbola.

Some navigation systems used by ships are based on the properties of hyperbolas. A lamp casts a hyperbolic shadow on a wall, and many telescopes use hyperboliclenses.

A hyperbola is a graph consisting of two branches. The hyperbolas we will consider will have branches that open either in the x-direction or in the y-direction.

Notice how the branches of the hyperbolaget closer to the dotted lines as the branchescontinue indefinitely. These dotted lines are calledAsymptotes.

GraphExample 4

1

9

2

4

3 22

yx

SolutionHow do we know that this is a hyperbola and not an ellipse? It is a hyperbola because there is subtraction sign between the two quantities on the left. If it were addition, it would be an ellipse.

Standard form is

12

2

2

2

b

ky

a

hx

Our equation is 19

)2(

4

)3( 22

yx

What is h? What is k?

What is a? What is b?

h = 3

a= 39 b = 24

k = – 2

The center is (3, -2).2

2

( )Since the quantity is the positive quantity, the branches of

the parabola will open in the -direction.

x h

ax

Use the center, ( 3, -2 ), a = 2, b = 3 to draw a reference rectangle. The diagonalsof this rectangle are the asymptotes of the hyperbola.

First, plot the center (3, -2). Since a = 2 and a2 is under the squared quantity containingthe x, move 2 units each way in the x-direction from the center. These are twopoints on the rectangle.Since b = 3 and b2 is under the squared quantity containing the y, move 3 unitseach way in the y-direction from the center. These are two more points on the rectangle. Draw the rectangle containing these four points, then draw the diagonalsof the rectangle as dotted lines. These are the asymptotes of the hyperbola.Sketch the branches of the hyperbola opening in the x-direction with the branchesapproaching the asymptotes. diagonals

GraphExample 5

1

254

22

xy

Standard form is

12

2

2

2

a

kx

b

hy

Our equation is 125

)(

4

)( 22

xy

What is h? What is k?

What is a? What is b?

h = 0

a= 24 b = 525

k = 0

The center is (0, 0).

Solution

GraphExample 6

99 22 xySolutionThis is a hyperbola since there is a subtraction sign between the two terms. Since the standard form for the equation of the hyperbola has a 1 on one side of the equal sign, divide both side of y2 – 9x2 = 9 by 9 to obtain a 1 on the right.

99 22 xy

9

9

9

9

9

22

xy

19

22

xy

The center is (0, 0).

direction.-y in theopen willparabola the

of branches thequantity, positive theis 9

quantity theSince2y

Divide both sides by 9.

Simplify.

h = 0

a= 39 b = 11

k = 0