mth 112 complete conic sectionscf.linnbenton.edu/mathsci/math/maurerv/upload/mth 112...and ( −2,...

1

© 2007 Linn-Benton Community College MTH 112 Conics

MTH 112 CONIC SECTIONS

The four basic types of conic sections we will discuss are: circles, parabolas, ellipses, and

hyperbolas. They were named “conic” by the Greeks who used them to describe the intersection of a

plane with a double cone. Although you may have seen equations for some of these curves in the

past, this time we will be approaching them from a different point of view.

Figures adapted from Mathworld.wolfram.com and vectosite.net.

I. CIRCLES

A circle is the set of points in a plane equidistant from a fixed point, called the center. The

distance from any point on the circle to the center is called the radius. If the center is C(h, k), and

the radius is r, then the distance from C to any other point, (x, y) on the circle is found by using the

distance formula.

rkyhx =−+− 22 )()( If we square both sides of this equation we get

222 )()( rkyhx =−+−

Which is the standard equation of a circle we have used in the past.

Example 1: Find the standard equation of a circle with radius 2 and center (4, −3). Then sketch

the graph.

Solution: We are given r = 2, h = 4, and k = −3,

so the equation of the circle is 2 2( 4) ( 3) 4x y− + + = .

X

Y

2 4 6 8

-6

-4

-2

2

0

2


If we multiply the standard form of a circle out and clear out the fractions, we will get an equation of

the general quadratic form: Ax2 + By

2 + Cx + Dy + E = 0. If A = B, this will be a circle.

Example 2: Put the following equation of a circle in standard form. State the center and radius.

2 22 2 8 4 22 0x y x y+ + − − =

Solution: Since there are coefficients of the squared terms that are not 1, we will start by

dividing each term in the equation by 2.

2 2

2 2

2 2 8 4 22 0

2 2 2 2 2 2

4 2 11 0

x y x y

x y x y

+ + − − =

+ + − − =

Now we rearrange the terms and complete the square on x and on y.

2 2

2 2

2 2

4 2 11

4 4 2 1 11 4 1

( 2) ( 1) 16

x x y y

x x y y

x y

+ + − =

+ + + − + = + +

+ + − =

The equation is in standard form. The center is ( 2,1)− , and the radius is r = 4.

Example 3: Write the equation of the circle with endpoints of the diameter at ( 2,1)− and (4, 3)− .

Solution: The center of the circle must be the midpoint of the diameter. To find it we will put

the diameter endpoints into the midpoint formula 1 2 1 2,2 2

x x y y+ +

to get

( )1 32 4 2 2, , (1, 1)

2 2 2 2

+ − − + − = = −

. This is the center of the circle and will be

substituted for ( , )h k in the standard form for a circle. Let h = 1 and k = −1.

2 2 2

2 2 2

2 2 2

( ) ( )

( 1) ( ( 1))

( 1) ( 1)

x h y k r

x y r

x y r

− + − =

− + − − =

− + + =

To get the radius, r, we note that the circle must pass through both of the endpoints on

the diameter. We can use one of them and substitute the values for x and y in the

equation and solve for r. Use the point ( 2,1)− . Let x = −2 and y = 1.

2 2 2

2 2 2

2

2

( 2 1) (1 1)

( 3) (2)

9 4

13

13

r

r

r

r

r

− − + + =

− + =

+ =

=

=

Now we can write the standard form of the equation of this circle.

2 2( 1) ( 1) 13x y− + + =

3


EXERCISES:

1. Find the equation of the circle with radius

9 and center (−3, 1).

2. Find the equation of the circle with radius

7 and center (4, −2).

3. a. Draw the set of points which are 5

units from (2, −3).

b. Find an equation for the set of points.

4. a. Draw the set of points which are 7

units from (0, 0).

b. Find an equation for the set of points.

For Exercises 5 and 6, identify the center and

radius of the given circles.

5. 2 2( 8) ( 11) 12x y+ + − =

6. 2 2( 6) ( 5) 18x y− + − =

For Exercises 7 and 8, identify the center and

radius, and sketch the graph of the circle. 7. 2 2( 2) ( 3) 49x y+ + − =

8. 2 2( 6) 36x y− + =

For Exercises 9 through 12, find the center

and radius, and then sketch the graph of each

circle.

9. 2 2 8 14 0x y y+ + + =

10. 2 2 14 24 0x y y+ − + =

11. 2 2 6 2 1 0x y x y+ − − + =

12. 2 2 4 6 4 0x y x y+ + − + =

For Exercises 13 through 16, find the equation

of a circle that satisfies the given information.

13. The diameter of the circle passes through

the points (−5, 6) and (7, −4).

14. The diameter of the circle passes through

the points (−3, −2) and (3, 4).

15. The center of a circle is at (7, −5) and a

point on the circle is (−2, 3).

16. The center of a circle is at (0, −8) and a

point on the circle is (9, 0)

4


II. PARABOLAS

A parabola is defined as the set of all points P(x, y) in a plane which are equidistant from a

fixed line called the directrix and a fixed point (not on the line) called the focus. If the vertex of a

parabola opening up or down is at (0,0), the focus is at (0, p), then the directrix is y = − p.

If P(x, y) is any point on the parabola. Then, by the definition of a parabola given above,

P(x, y) is equidistant from both the focus (0, p) and the directrix, y = − p. The point on the directrix

closest to (x, y) is (x, − p). Let d1 be the distance from P(x, y) to (0, p) and let d2 be the distance from

P(x, y) to (x, −p) as shown in the next figure.

By the definition of a parabola, d1 = d2. Using the distance formula, we have 2 2 2 2

2 2 2 2 2 2

2 2 2 2 2

( 0) ( ) ( ) ( ( )) or

2 0 2 Simplify and square both sides

2 2

x y p x x y p

x y py p y py p

x y py p y py p

− + − = − + − −

+ − + = + + +

+ − + = + +

Now subtracting y2, p

2, and adding 2py to both sides leaves 2 4x py= .

This is the standard form of a parabola that opens up or down and has its vertex at (0, 0), focus at

(0, p), and directrix, y = − p. Recall that the axis of symmetry for a parabola passes through the

vertex. Notice that the focus is on the axis of symmetry for the parabola and the directrix is

perpendicular to the axis of symmetry.

X

Y

P(x, y) • focus

(0, p)

y p= −

•

directrix

X

Y

P(x, y) • focus

(0, p)

y p= −

•

directrix •

(x, − p)

d1

d2

5


X

Y

-2 2

-2

2

4

0

• (0, 0.5)

y = −0.5

focus

• •

Example 1: State the coordinates of the vertex and the focus, and the equation of the directrix of

the parabola 2 2x y= . Then sketch the graph.

Solution: First find p. Compare the standard form of a parabola to the given equation and note

that 4p = 2, so p = 1

2 = 0.5. The vertex is (0, 0), so the focus is p = 0.5 units up along

the axis of symmetry at (0, 0.5), and the directrix is y = −0.5. Because p is positive,

this parabola opens up. To determine how wide to sketch the parabola, select a

y-value and use the equation to find the x-values that are paired with it. Select y = 2,

for example, and substitute y = 2 into 2 2x y= , then solve for x.

2

2

2(2)

4

4

2

x

x

x

x

=

=

= ±

= ±

Plot the points (2, 2)

and (−2, 2) and complete

the sketch of the parabola.

Example 2: State the coordinates of the vertex and the focus, and the equation of the directrix of

the parabola 2 12x y= − . Then sketch the graph.

Solution: Find p: 4p = −12, so p = −3. Because p is negative, this parabola opens down.

The vertex is (0, 0), focus is (0, −3), and directrix is y = 3. Plot two additional points

by selecting a y-value, say y = −3, substitute it into the equation of the parabola and

solve for x.

If y = −3, then ( )2 12 3 36x = − − = , and 6x = ± . Plot (6, −3) and (−6, −3).

Standard Form of a Parabola:

Vertex at (0, 0), focus at (0, p), and directrix at y = −p.

2

4x py=

If p > 0, the parabola opens up. If p < 0, the parabola opens down.

6


X

Y

-8 -6 -4 -2 2 4 6 8

-4

-3

-2

-1

1

2

3

4

0

y = 3

•

focus

(0, −3)

• •

We have seen that a parabola with general form, 2 4x py= , opens up if 0p > and opens

down if 0p < . A parabola can also open left or right. A parabola with vertex at (0, 0) which opens

left or right has the general form: 2 4y px= .

If 0p > , the parabola opens right as shown. If 0p < , the parabola opens left as shown.

X

Y

focus

(p, 0)

•

directrix

x = −p

•

directrix

x = p

focus

(−p, 0)

Standard Form of a Parabola:

Vertex at (0, 0), focus at (p, 0), and directrix at y = −p.

2

4y px=

If p > 0, the parabola opens right. If p < 0, the parabola opens left.

7


X

Y

-6 -4 -2 2

2

4

6

8

0

• • •

y = 1

F

•

V

The standard form of a parabola with vertex at (h, k) can be found by translating the forms we have

seen already from vertex (0, 0) to vertex ( , )h k by substituting x h− for x and y k− for y to get:

Opens up or down: (x – h)2 = 4p(y – k) with focus F(h + p, k) and directrix y = − p.

Opens right or left: (y – k)2 = 4p(x – h) with focus F(h, k + p) and directrix x = − p.

Example 3: State the coordinates of the vertex and focus, and the equation of the directrix, then

sketch the graph of the parabola (x + 2)2 = 8(y – 3).

Solution: 4p = 8 so p = 2. The vertex is at V(−2, 3) and the focus is at F(−2, 3 + 2) or F(−2, 5).

The directrix is the line y = 3 − 2, or y = 1. Plot two additional points. If y = 5, then

2

2

( 2) 8(5 3)

( 2) 16

2 4

2 4

6 or 2

x

x

x

x

x

+ = −

+ =

+ = ±

= − ±

= −

Plot the points (2, 5) and (−6, 5).

In Example 3, we used the y-coordinate of the focus to generate two points on the parabola. Those

points lie to the right and to the left the focus. The line segment joining those two points that passes

through the focus is called the latus rectum.

Example 4: State the coordinates of the vertex and focus, and the equation of the directrix, then

sketch the graph of the parabola (y – 1)2 = −4(x – 2). Find the endpoints of the latus

rectum.

Solution: 4p = − 4 so p = −1. Because p < 0,

this parabola opens left. The vertex

is V(2, 1). The focus is F(2 − 1, 1)

or F(1, 1). The directrix is the line

x = 2 −(−1), or x =3. When x = 1, 2( 1) 4(1 2)y − = − − , so y = 3 or −1.

Plot (1, 3) and (1, −1). These are

the endpoints of the latus rectum.

The latus rectum is shown as a

dashed segment in the figure.

X

Y

-2 2 4

-2

2

4

0

x = 3

• F

•

•

• V

8


Parabolas with axes of symmetry parallel to a coordinate axis and vertex (h, k)

p Focus Directrix Equation Description

p > 0 (h + p, k) x = h− p (y − k)2 = 4p(x− h) Axis of symmetry parallel to x-axis,

opens right

p < 0 (h + p, k) x = h− p (y − k)2 = 4p(x − h) Axis of symmetry parallel to x-axis,

opens left

p > 0 (h, k + p) y = k− p (x − h)2 = 4p(y − k) Axis of symmetry parallel to y-axis,

opens up

p < 0 (h, k + p) y = k− p (x − h)2 = 4p(y − k) Axis of symmetry parallel to y-axis,

opens down

If the equation of a parabola is put into the general quadratic form with integer coefficients it will be

of the form Ax2 + Bx + Cy + D = 0 or Ay

2 + Bx + Cy + D = 0. Notice that the equations do not

have both x2 and y

2 terms. Using completing the square the equations can be put into the standard

form for a parabola.

EXERCISES:

For Exercises 1 through 8 state the coordinates

of the vertex and focus, the equation of the

directrix, and sketch the graph. Also find and

label the endpoints of the latus rectum.

1. y2 = 4x

2. y2 = −6x

3. x2 = −8y

4. x2 = 16y

5. (y – 1)2 = 20(x + 2)

6. (y − 3)2 = −12(x + 1)

7. (x − 4)2 = 4(y + 3)

8. (x – 5)2 = −4(y – 2)

9. Sketch all the points that are equidistant

from the line x = −2 and the point (2, 3).

Find an equation for this curve.

10. Sketch all the points that are equidistant

from the line y = 6 and the point (−1, 1).

Find an equation for this curve.

11. Find the equation of the parabola with the

focus at (2, 5) and directrix y = −1.

12. Find the equation of the parabola with

focus at (−4, −2) and directrix x = −3.

Exercises 13 and 14: The surface formed by

rotating a parabola about its axis of symmetry

is called a paraboloid of revolution. If the

paraboloid of revolution is sliced through its

axis of symmetry, the cross section is a

parabola.

13. The reflective surface of a good flashlight

is in the shape of a paraboloid of

revolution. Any light radiating from a

light bulb set at the focus will reflect off

the mirrored surface as parallel beams of

light. (See the figure.) If the light bulb is

set 1.2 cm from the base of the surface,

find the equation of the parabolic cross

section. Hint: Place the vertex at (0, 0).

How wide is the flashlight at its widest

point if it is 4 cm deep at the center?

Round to the nearest tenth of a centimeter.

14. A satellite dish is in the shape of a

paraboloid of revolution. Signals that hit

the surface of the satellite dish are

reflected to one point, the focus, where the

receiver is located. At its widest point, the

satellite dish is 10 feet wide. It is 3.5 feet

deep at its center. Determine the where

the receiver should be placed. Hint: Find

Focus

9


the equation of the parabolic cross section

and assume the base of the satellite dish is

at (0, 0).

15. An arched doorway in the shape of a

parabola is 12 feet at its tallest point. It is 4

feet wide at the base. How wide is the

doorway at a height of 6 feet above the

floor? Round to the nearest tenth of a

foot. Hint: Find an equation of the

parabola first.

16. A bridge built in 1914 was constructed

with a “parabolic hinged-rib arch.” The

arch is 58 feet across at its widest point,

where it intersects the concrete walls at a

point 10 feet above the sidewalk. The

height from the pavement to the highest

part of the arch is 35 feet. How wide is

the arch 15 feet above the pavement?

Round to the nearest tenth of a foot. Hint:

Find an equation of the parabola first.

For Problems 17 and 18, identify whether the

equation represents a circle, parabola or

neither. If it is a circle or parabola, put it into

standard form.

17. a. 2 2 4 6 3 0x y x y+ + − − =

b. 2 8 6 28 0x x y+ + + =

c. 2 22 4 6 0x y x− + + − =

18. a. 2 2 6 2 5 0x y x y− + + − =

b. 2 8 4 12 0y x y+ + − =

c. 2 22 2 12 6 0x y x+ + − =

10


X

Y

• F1 (−c, 0)

• F2 (c, 0)

(0, −b)

•

V2 (a, 0)

• V1 (−a, 0)

•

(0, b) •

III. ELLIPSES

An ellipse is the set of points in a plane, the sum of whose distances from two fixed points,

called the foci, is a constant. If we let the sum of the two distances, F1P + F2P = 2a where a is a

constant, then the ellipse with center at the origin, longest axis (called the major axis) along the x-

axis, shorter axis (minor axis) along the y-axis has the following standard equation.

2 2

2 21

x y

a b+ = where 2 2 2

a b c− =

Note: The relationship 2 2 2a b c− = does not match the Pythagorean theorem. And that in this

equation 2 2a b> . The graph of an ellipse that matches this equation will have the following general

form. The points at the ends of the major axes are called vertices, in the graph these are labeled V1

and V2.

If the center is at the origin, and the two foci are on the y-axis, the major axis will be vertical and the

minor axis will be horizontal. Then the standard equation will be 2 2

2 21

y x

a b+ = where 2 2 2

a b c− =

And a graph might look like

X

Y

• F2 (0, −c)

• F1 (0, c)

V1 (0, a) •

• V2 (0, −a)

(b, 0) •

(−b, 0) •

11


Note: You can tell whether the major axis of an ellipse is horizontal or vertical by comparing the

values underneath the x2 and y

2 terms. If the largest value is below x

2, the major axis is horizontal.

If the largest value is below y2, the major axis is vertical.

Example 1: Find the coordinates of the vertices and foci and sketch: 5x2 + 3y

2 = 15.

Solution: We have to put this into the standard form 2 2

2 21

x y

a b+ = or

2 2

2 21

y x

a b+ = first. Create a

1 on the right-hand side of the equation by dividing each term in the equation by 15.

2 25 3 15

15 15 15

x y+ = , which simplifies to

2 2

13 5

x y+ = .

Since 5 > 3, and 5 is the denominator of the y2 term, the major axis is vertical.

a2 = 5, so 5a = ; and b

2 = 3, so 3b = . Since a

2 – b

2 = c

2, c

2 = 5 – 3 and c = 2 .

The center is C(0, 0). The vertices and foci are equidistant from the center. The

vertices occur at V1 ( )0, 5 and V2 ( )0, 5− . The foci occur at F1 ( )0, 2 and

F2 ( )0, 2− . Also note that the endpoints of the minor axis are at ( )3,0 and

( )3,0− , however, these points are not called vertices. They are used for graphing

purposes only.

Example 2: Find the equation of an ellipse, centered at the origin with V1 ( 6,0)− and one focus at

(2, 0). Also give the coordinates of the other vertex and focus.

Solution: Notice that the given vertex and focus points are both on the x-axis, so the major axis

is horizontal. Thus, a2 must be the denominator of the x

2 term in the equation. The

value of a is defined as the distance from the center to a vertex point. The distance

from the coordinates of V1 to the origin is 6 units, so a = 6. The value of c is defined

as the distance from the center to a focus point. The distance from (2, 0) to the origin

is 2 units, so c = 2. Since a2 – b

2 = c

2, we have 36 – b

2 = 4, so b

2 = 36 – 4 = 32. The

equation is 2 2

136 32

x y+ = . Since the vertex points and the focus points are equidistant

from the center, the other vertex is at V2(6, 0) and the other focus is at (−2, 0).

X

Y

-2 -1 1 2

-3

-2

-1

1

2

3

0

V1

F1

•

V2

•

F2

12


To translate the center of an ellipse from (0, 0) to any point ( , )h k we substitute x h− for x and y k−

for y to get:

Example 3: Find the coordinates of the center, vertices and foci for 2 2( 2) ( 3)

19 4

x y− ++ = , and

sketch the graph.

Solution: The center is C(2, −3). The major axis is horizontal since the larger number is in the

denominator of the x term. Thus, a2 = 9 and a = 3; b

2 = 4 and b = 2. Find the value

for c by using the values of a and b: c2 = a

2 – b

2 = 9 – 4 = 5, so c = 5 .

The vertices are found by adding or subtracting the value of a from the x-coordinate

of the center. Thus, the vertices are (2 ± 3, −3), or V1(−1, −3) and V2(5, −3).

Similarly, the foci are (2 ± 5 , −3), or F1(2 − 5 , −3) and F2(2 + 5 , −3). To

graph the ellipse, locate the endpoints of the minor axis at (2, −3 ± 2), or (2, −5) and

(2, −1).

Standard Form of an Ellipse

Center (h, k), 2 2a b>

2 2

2 2

( ) ( )1

x h y k

a b

− −+ = Major Axis is Horizontal

2 2

2 2

( ) ( )1

y k x h

a b

− −+ = Major Axis is Vertical

X

Y

-2 -1 1 2 3 4 5 6

-6

-5

-4

-3

-2

-1

1

0

V1 V2 • • F1 F2 C(2, −3)

13


Example 4: Find the equation of an ellipse with center at ( 4,2)− and major axis parallel to the y-

axis, length of major axis equal to 8, and length of minor axis equal to 6.

Solution: Since the length of the major axis is 2a, a = 4. The length of the minor axis is 2b, so

b = 3. The desired ellipse has the following equation.

2 2( 2) ( 4)

116 9

y x− ++ =

An ellipse has a very special reflective property. If a source of sound is placed at one focus of the

ellipse, then the sound wave will reflect off the ellipse and travel to the other focus. The interior of

the dome of St. Paul’s Cathedral in London is built with elliptical ceilings, so a whisper uttered at

one focus can be clearly heard at the other. A room built in such a way is called a Whispering

Gallery. If the ellipse were made up of a reflective surface, then light rays emitted from one focus

would reflect to the other focus. In medicine, a machine constructed in the shape of a half ellipsoid

produces shock waves from one focus that can break up a kidney stone at the other focus, without

surgery.

Example 5: A pool table will be constructed in the shape of an ellipse,

50 inches wide and 100 inches long. How far from the

center should the cue ball and pocket be placed, so that

when the cue ball is hit, it will carom off the wall and land

in the pocket?

Solution: The major axis of the ellipse is 100 inches long and the minor axis is 50 inches long,

thus a = 50 and b = 25. If we imagine placing the center of the ellipse at the origin of

a coordinate system, with the major axis horizontal, then we can find the locations of

the foci, which are where the cue ball and pocket must be placed to complete the

desired shot. Find c: 2 2 2 2 250 25 1875c a b= − = − = , thus 25 3c = ≈ 43.3 inches.

Place the cue ball and pocket along the major axis 43.3 inches to the right and left of

the center.

X

Y

-60 -50 -40 -30 -20 -10 10 20 30 40 50 60

-30

-20

-10

10

20

30

0• •

cue ball pocket

14


Ellipse with center at (h, k) and major axis horizontal or vertical

Center Major axis Foci Vertices Equation

(h, k) Horizontal (h + c, k)

(h – c, k)

(h + a, k)

(h – a, k)

2 2

2 2

( ) ( )1

x h y k

a b

− −+ =

a > b, a2 – b

2 = c

2

(h, k) Vertical (h, k + c)

(h, k – c)

(h, k + a)

(h, k – a)

2 2

2 2

( ) ( )1

y k x h

a b

− −+ =

a > b, a2 – b

2 = c

2

So far we have seen the equation of an ellipse given in standard form. But any equation of the form 2 2 0Ax By Cx Dy E+ + + + = in which A ≠ B and A and B have the same sign will be an ellipse.

Example 6: Find the coordinates of the center and orientation of the major axis for the ellipse

given by 2 225 4 50 24 339 0x y x y+ + − − = .

Solution: We need to complete the square on the variables x and y to put the equation in

standard form. To set the equation up, we will factor the coefficient of 2x from the x-

terms and the coefficient of 2y from the y-terms, then complete the square.

2 2

2 2

2 2

2 2

2 2

2 2

25 4 50 24 339 0

25( 2 ) 4( 6 ) 339

25( 2 1) 4( 6 9) 339 25 36

25( 1) 4( 3) 400

25( 1) 4( 3) 400

400 400 400

( 1) ( 3)1

16 25

x y x y

x x y y

x x y y

x y

x y

x y

+ + − − =

+ + − =

+ + + − + = + +

+ + − =

+ −+ =

+ −+ =

The center of the ellipse is the point ( 1,3)C − . And since the larger denominator is

under the y variable the major axis is vertical.

EXERCISES:

For Exercises 1 through 6, find the

coordinates of the center, vertices, and foci,

and sketch the graph.

1. 2 2

149 16

x y+ =

2. 2 2

19 64

x y+ =

3. 2 210 25 100x y+ =

4. 2 215 6 60x y+ =

5. 2 2( 1) ( 2)

111 36

x y− ++ =

6. 2 2( 3) ( 1)

149 13

x y− ++ =

15


For Exercises 7 through 14, find the equation

of the ellipse satisfying the given information.

7. Center at (0, 0), focus at (−1, 0), vertex at

(3, 0)

8. Center at (0, 0), focus at (0, 3), vertex at

(0, −7)

9. Center at (4, −5), length of horizontal

major axis is 10, length of minor axis is 6.

10. Center at (−2, 3), one focus at (−2, 0),

length of minor axis is 1

11. Foci at (0, ± 3), x-intercepts are (±2, 0)

12. Foci are at (2, 0) and (8, 0), length of

minor axis is 6

13. Foci at (± 2, 0), sum of the distances from

the foci to any point is 8

14. Foci at (0, ±5), sum of the distances from

the foci to any point is 12

15. Determine if the quadratic equation given

represents an ellipse. If it does, then

change it to standard form, state its center,

and whether the major axis is horizontal or

vertical.

a. 2 24 9 36 0x y+ − =

b. 2 24 8 4 8 0x y x y+ − + − =

c. 24 8 15 25 0x x y− + − =

16. Determine if the quadratic equation given

represents an ellipse. If it does, then

change it to standard form, state its center,

and whether the major axis is horizontal or

vertical.

a. 2 23 12 6 15 0x y x y+ − + − =

b. 2 26 5 150 0x y− − =

c. 2 29 64 90 128 287 0x y x y+ − + − =

Exercises 17 through 20: The orbits of the

planets in our solar system are elliptical with

the sun at one focus. Some orbits are nearly

circular, while others are more elongated, or

eccentric. Eccentricity, e, of an ellipse is

defined as c

a. For a circle, eccentricity is

zero. Eccentricities approach 1 as ellipses

grow longer and narrower.

17. At perihelion, the Earth is closest to the

sun, a distance of 147.5 million miles.

The Earth is farthest away from the sun at

aphelion, a distance of 152.6 million

miles. Both distances are measured along

the major axis of the ellipse as shown.

Find the eccentricity of the Earth’s orbit.

Round to the nearest thousandth.

18. The length of the major axis of Neptune’s

orbit is about 2,265,363,900 km and the

length of the minor axis is about

2,265,221,800 km. Find the eccentricity

of Neptune’s orbit. Round to the nearest

thousandth.

19. Mercury’s orbit is more eccentric,

e = 0.2056, than that of the Earth. At

aphelion, Mercury is 69.8 million miles

away from the sun. Find Mercury’s

perihelion distance. Round to the nearest

million miles. See Exercise 17 for

definitions.

20. The length of Mars’ minor axis is 227.92

million km. The eccentricity of the orbit

of Mars is 0.0935. Find the length of the

major axis of Mars’ orbit. Round to the

nearest million kilometers.

perihelion

aphelion

16


21. A brick wall has an opening in the shape

of an ellipse, as shown in the following

figure. At its widest points, the opening is

4 feet tall and 2.5 feet wide. You would

like to pass a rectangular box through the

window that measures 3.5 feet tall, 1.25

feet wide, and 5 feet deep. The opening in

the wall is tall enough to accommodate the

3.5 foot height, but is it wide enough to

allow the box through? Explain. Hint:

First find the equation of the ellipse by

placing the center of the ellipse at the

origin of a coordinate system.

22. A bridge is built so that the arch is in the

shape of a semi-ellipse. See the following

figures. If the major axis of the ellipse is

35 feet wide and the minor axis is 14 feet

tall, then how wide is the ellipse at a point

3 feet from the highest point in the arch?

Hint: First find the equation of the ellipse

by placing the center of the ellipse at the

origin of a coordinate system. Round to

the nearest tenth.

Semi-elliptical Arched Bridge

1.25′

3.5′

5′

17


IV. Hyperbolas

A hyperbola is the set of points in a plane such that the difference of the distances from two fixed

points, called foci, is a positive constant.

The line that joins the two foci, F1 and F2, is called the transverse axis. The vertices, V1 and V2, are

located where the hyperbola intersects the transverse axis. The center of the hyperbola is located

along the transverse axis halfway between the foci. A possible graph of a hyperbola with a

horizontal transverse axis is shown. In the figure P(x, y) represents any point on the hyperbola.

To create an equation for the hyperbola, the difference in the distances from F1 to P and F2 to P must

be some positive constant, but what is the constant? Notice that if we let P be one of the vertex

points, then the difference in the distances would exactly equal 2a. With 2a as the constant, we

will write down the difference in the distances, using the distance formula.

The distance from F1 to P minus the distance from F2 to P is 2a.

2 2 2 2( ( )) ( 0) ( ) ( 0) 2x c y x c y a− − + − − − + − =

( ) ( )

( )

2 2 2 2

2 22 2 2 2

2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2

2 2 2

2 2 2

22 2 2 2

2 2 2 4 2 2

( ) 2 ( )

( ) 2 ( )

( ) 4 4 ( ) ( )

2 4 4 ( ) 2

4 4 4 ( )

( )

( ) ( )

2 2

x c y a x c y

x c y a x c y

x c y a a x c y x c y

x cx c y a a x c y x cx c y

cx a a x c y

cx a a x c y

cx a a x c y

c x a cx a a x cx

+ + = + − +

+ + = + − +

+ + = + − + + − +

+ + + = + − + + − + +

− = − +

− = − +

− = − +

− + = − +( )2 2c y+

X

Y

• • F1(−c, 0) F2(c, 0)

• P(x, y)

• V1(−a, 0)

• V2(a, 0)

Transverse Axis Center

•

Isolate one radical expression

Square both sides

Square the binomials

Combine like terms

Divide each term by 4

Square both sides

Square the binomials

18


2 2 2 4 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2

2 2

( ) ( )

c x a cx a a x a cx a c a y

c a x a y a c a

− + = − + +

− − = −

To simplify the equation further, notice the quantity 2 2c a− occurs in two places. Refer to the graph

of the hyperbola and verify that c is the distance from the center to a focus and a is the distance from

the center to a vertex. From that relationship we note that c > a, so 2 2c a− must be a positive

number, let's call it 2b , and in the equation make the substitution 2 2 2

b c a= − .

2 2 2 2 2 2 2 2

2 2 2 2 2 2

2 2 2 2 2 2

2 2 2 2 2 2

2 2

2 2

( ) ( )

1

c a x a y a c a

b x a y a b

b x a y a b

a b a b a b

x y

a b

− − = −

− =

− =

− =

And this final equation is the standard form of the equation of a hyperbola with its center at (0, 0)

and a horizontal transverse axis.

Look back at the graph of the hyperbola and note that if 2 2 2b c a= − , then 2 2 2

c a b= + , and there is a

geometric meaning for b. Every hyperbola has a fundamental rectangle, which you can see sketched

in the following figure.

The rectangle is such that its extended diagonals are the asymptotes of the hyperbola. The rectangle

has length 2a and width 2b. When graphing a hyperbola, be sure to sketch the fundamental

rectangle and its diagonals and use them for asymptotes for the hyperbola.

The hyperbola we have looked at so far has its center at the origin, (0, 0). To translate the center to

any other point (h, k) in the xy-plane we substitute x h− for x and y k− for y to get the following.

Distribute

Collect like terms and factor

Substitute 2 2 2b c a= −

Divide each term by 2 2a b

X

Y

• • F1(−c, 0) F2(c, 0)

• V1(−a, 0)

• V2(a, 0)

(0, −b)

• (0, b)

•

19


The Standard Form of a Hyperbola with a Horizontal Transverse Axis

2 2

2 2

( ) ( )1

x h y k

a b

− −− =

The center of the hyperbola is ( , )h k .

The foci are ( , )h c k± , where 2 2 2c a b= + .

The vertices are ( , )h a k± .

A hyperbola can have a vertical transverse axis also as shown next.

The equation for a hyperbola with a vertical transverse axis is as follows.

The Standard Form of a Hyperbola with a Vertical Transverse Axis

2 2

2 2

( ) ( )1

y k x h

a b

− −− =

The center of the hyperbola is ( , )h k .

The foci are ( , )h k c± , where 2 2 2c a b= + .

The vertices are ( , )h k a± .

Example 1: Sketch the graph of the hyperbola. State the coordinates of the center, the vertices

and the foci. Find the equations of the asymptotes. 2 2( 2) ( 1)

14 16

y x+ −− =

Solution: The center is (1, 2)− . The hyperbola has a vertical transverse axis. We will draw the

fundamental rectangle using a = 2, and b = 4. Then sketch the asymptotes and the

hyperbola opening up and down. The vertices occur at (1, −2 ± 2), or 1(1, 0)V and

2 (1, 4)V − . To get the foci we need to use 2 2 2c a b= + , so 2 2 2(2) (4) 20c = + = and

X

YTransverse Axis

20


20 2 5c = = . The coordinates of the foci are (1, 2 2 5)− ± , or F1(1, −2 + 2 5 )

and F2(1, −2 − 2 5 ).

To find the equations of the asymptotes, notice that each asymptote passes through

the center, (1, 2)− , and a corner of the fundamental rectangle. The positively-sloped

asymptote has a rise of 2 and a run of 4, for a slope of 2 1

4 2= . Similarly, the

negatively-sloped asymptote has a slope of 2 1

4 2

− −= . Using the point-slope form of

the equation of a line we obtain the equations of the asymptotes.

1 1 1 1( ) ( )

1 1( 2) ( 1) ( 2) ( 1)

2 2

3 5

2 2 2 2

y y m x x y y m x x

y x y x

x xy y

− = − − = −

−− − = − − − = −

−= − = −

You may have encountered hyperbolas without realizing it. A light bulb in a cylindrical lampshade

will cast light on the wall in the shape of a hyperbola. Two pebbles dropped into still water will

produce waves in concentric circles, the intersections of which form hyperbolas. Nuclear reactors

must have cooling towers that can withstand high winds and be made of as little material as possible.

The most efficient shape has a hyperbolic cross section.

X

Y

-4 -2 2 4 6

-8

-6

-4

-2

2

4

0•

•

•

•

•

F1

F2

V1

V2

C

Lamp Light Intersecting Waves Cooling Tower

21


Example 2: Complete the square to put the equation for the hyperbola in standard form. Then

state the center and direction of the transverse axis.

2 22 9 4 54 97 0x y x y− + − − = .

Solution: Rearrange the terms to gather x-terms together and y-terms together. Factor so that

the leading coefficient on each squared term is 1. 2 2

2 2

2 2

2 9 4 54 97 0

2( 2 ) 9( 6 ) 97

2( 2 1) 9( 6 9) 97 2 81

x y x y

x x y y

x x y y

− + − − =

+ − + =

+ + − + + = + −

2 2

2 2

2 2

2( 1) 9( 3) 18

2( 1) 9( 3) 18

18 18 18

( 1) ( 3)1

9 2

x y

x y

x y

+ − + =

+ +− =

+ +− =

The center of the hyperbola is ( 1, 3)− − and the transverse axis is horizontal.

Example 3: A cooling tower with a cross section in the shape of a

hyperbola is 500 feet tall. Its narrowest point is 375 feet

above the ground. If the hyperbola is placed in a coordinate

system, with its base centered along the x-axis, then the

hyperbola has the following equation.

( )

22 3751

9025 63504

yx −− =

How wide is the tower at its base?

Solution: Consider the graph of the hyperbola. The

center is at (0, 375). To determine the width

at the base, find the x-coordinates when

y = 0. The distance between those two

points is the width. Substitute y = 0 into the

equation and solve for x.

( )22

2 2

2

0 3751

9025 63504

3751

9025 63504

3759025 1

63504

170.32

x

x

x

x

−− =

= +

= ± +

≈ ±

The endpoints of the base occur at (−170.32, 0) and (170.32, 0), so the base has a

width of 2(170.32) = 340.64 feet.

X

Y

-200 -100 100 200

100

200

300

400

500

0

22


EXERCISES:

Exercises 1 through 6: Sketch a graph of each

of the following hyperbolas, with the

fundamental rectangle and asymptotes. State

the coordinates of the center, vertices and foci,

and give the equations of the asymptotes.

1. 2 2

19 4

x y− =

2. 2 2

116 25

y x− =

3. 2 2( 4) ( 1)

19 36

y x+ +− =

4. 2 2( 2) ( 3)

14 9

x y− +− =

5. 2

2( 2)( 1) 1

49

yx

−− − =

6. 2

2 ( 3)( 1) 1

4

yx

−+ − =

Exercises 7 through 10: Complete the square

to put the equation for the hyperbola in

standard form. State the coordinates of the

center and the direction of the transverse axis.

7. 2 29 36 6 18 0x y x y− − − + =

8. 2 29 36 72 0y x x− + − =

9. 2 23 2 6 12 27 0x y x y− − − − =

10. 2 29 2 54 62 0y x x y− + + + =

11. The Kobe Port Tower is a hyperbolic

telecommunications tower in Kobe, Japan.

It has a cross section that is a hyperbola.

The height of the tower is 108 meters and

there is an observation deck at a height of

90.28 meters. If an equation of the cross

section of the tower is 2 2( 65.5)

116.81 515.68

x y −− = , then find the

diameter of the observation deck. Round

to the nearest tenth.

12. Find the diameter of the base of the Kobe

Port Tower from Exercise 11. Round to

the nearest tenth.

13. A cooling tower, like the one pictured in

Example 3, is 450 feet tall. The narrowest

point is 390 feet above the ground, where

it is 180 feet wide. The base is 320 feet

wide. Find the diameter of the top of the

tower. Round to the nearest tenth.

14. A cooling tower, like the one pictured in

Example 3, is 300 feet tall. The narrowest

point is 265 feet above the ground, where

it is 170 feet wide. The diameter at the top

of the tower is 190 feet. Find the diameter

of the base of the tower. Round to the

nearest tenth.

Exercises 15 through 20: Classify the graph

of the equation as a circle, a parabola, an

ellipse or a hyperbola.

15. 2 24 6 16 21 0x y x y+ − + + =

16. 2 24 4 3 0x y x− − − =

17. 2 4 4 0y y x− − =

18. 225 10 200 119 0x x y− − − =

19. 2 29 9 36 6 34 0x y x y+ − + + =

20. 2 ( ) (3 2 2 )x x y y y x− = − −

Kobe Port Tower

mth 112 complete conic sectionscf.linnbenton.edu/mathsci/math/maurerv/upload/mth 112...and ( −2,...

Documents