Chapter 9: Conics

Section 9.1 Ellipses ..................................................................................................... 579

Section 9.2 Hyperbolas ............................................................................................... 598
Section 9.3 Parabolas and Non-Linear Systems ......................................................... 618
Section 9.4 Conics in Polar Coordinates..................................................................... 631

In this chapter, we will explore a set of shapes defined by a common characteristic: they
can all be formed by slicing a cone with a plane. These families of curves have a broad
range of applications in physics and astronomy, from describing the shape of your car
headlight reflectors to describing the orbits of planets and comets.

Section 9.1 Ellipses

The National Statuary Hall1 in Washington, D.C. is an
oval-shaped room called a whispering chamber because
the shape makes it possible for sound to reflect from the
walls in a special way. Two people standing in specific
places are able to hear each other whispering even
though they are far apart. To determine where they
should stand, we will need to better understand ellipses.

An ellipse is a type of conic section, a shape resulting from intersecting a plane with a
cone and looking at the curve where they intersect. They were discovered by the Greek
mathematician Menaechmus over two millennia ago.

The figure below2 shows two types of conic sections. When a plane is perpendicular to
the axis of the cone, the shape of the intersection is a circle. A slightly titled plane
creates an oval-shaped conic section called an ellipse.

1
Photo by Gary Palmer, Flickr, CC-BY, https://www.flickr.com/photos/gregpalmer/2157517950
2
Pbroks13 (https://commons.wikimedia.org/wiki/File:Conic_sections_with_plane.svg), “Conic sections
with plane”, cropped to show only ellipse and circle by L Michaels, CC BY 3.0
This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2020.
This material is licensed under a Creative Commons CC-BY-SA license. This chapter contains content
remixed from work by Lara Michaels and work from OpenStax Precalculus (OpenStax.org), CC-BY 3.0.
580 Chapter 9

An ellipse can be drawn by placing two thumbtacks in a piece of

cardboard then cutting a piece of string longer than the distance
between the thumbtacks. Tack each end of the string to the
cardboard, and trace a curve with a pencil held taught against
the string. An ellipse is the set of all points where the sum of
the distances from two fixed points is constant. The length of
the string is the constant, and the two thumbtacks are the fixed
points, called foci.

Ellipse Definition and Vocabulary

An ellipse is the set of all points Q(x, y ) for which the sum of the distance to two
fixed points F1 (x1 , y1 ) and F2 (x2 , y2 ) , called the foci (plural of focus), is a constant k:
d (Q, F1 ) + d (Q, F2 ) = k .
y
Q

d(Q,F1)
d(Q,F2)
x
F2 F1

The major axis is the line passing through the foci.

Vertices are the points on the ellipse which intersect the major axis.
The major axis length is the length of the line segment between the vertices.
The center is the midpoint between the vertices (or the midpoint between the foci).
The minor axis is the line perpendicular to the minor axis passing through the center.
Minor axis endpoints are the points on the ellipse which intersect the minor axis.
The minor axis endpoints are also sometimes called co-vertices.
The minor axis length is the length of the line segment between minor axis endpoints.
y

Note that which axis is major and which is minor Major axis
will depend on the orientation of the ellipse. In Vertices
the ellipse shown at right, the foci lie on the y Minor axis
axis, so that is the major axis, and the x axis is Foci endpoints
the minor axis. Because of this, the vertices are
the endpoints of the ellipse on the y axis, and the x
minor axis endpoints (co-vertices) are the
endpoints on the x axis.
Minor axis
 Section 9.1 Ellipses 581

Ellipses Centered at the Origin

From the definition above we can find an equation for an ellipse. We will find it for a
ellipse centered at the origin C(0,0) with foci at F1 (c,0) and F2 (− c,0) where c > 0.

Suppose Q(x, y ) is some point on the ellipse. The distance from F1 to Q is

d (Q, F1 ) = (x − c )2 + ( y − 0)2 = (x − c )2 + y 2
Likewise, the distance from F2 to Q is
d (Q, F2 ) = (x − (− c ))2 + ( y − 0)2 = (x + c )2 + y 2
From the definition of the ellipse, the sum of these distances should be constant:
d (Q, F1 ) + d (Q, F2 ) = k so that
(x − c )2 + y 2 + (x + c )2 + y 2 =k

If we label one of the vertices (a,0) , it should satisfy the equation above since it is a
point on the ellipse. This allows us to write k in terms of a.
(a − c )2 + 02 + (a + c )2 + 02 =k
a−c + a+c = k Since a > c, these will be positive
( a − c) + (a + c) = k
2a = k

Substituting that into our equation, we will now try to rewrite the equation in a friendlier
form.
(x − c )2 + y 2 + (x + c )2 + y 2 = 2a Move one radical
(x − c )2 + y 2 = 2a − ( x + c ) + y 2
2
Square both sides
2 2
 (x − c )2 + y 2  =  2a − (x + c )2 + y 2  Expand
   
(x − c )2 + y 2 = 4a 2 − 4a (x + c )2 + y 2 + (x + c )2 + y 2 Expand more
x 2 − 2 xc + c 2 + y 2 = 4a 2 − 4a (x + c ) + y 2 + x 2 + 2 xc + c 2 + y 2
2

Combining like terms and isolating the radical leaves

4a (x + c )2 + y 2 = 4a 2 + 4 xc Divide by 4
a (x + c )2 + y 2 = a 2 + xc Square both sides again
a ((x + c) + y ) = a + 2a xc + x c
2 2 2 4 2 2 2
Expand
a (x + 2 xc + c + y ) = a + 2a xc + x c
2 2 2 2 4 2 2 2
Distribute
582 Chapter 9

a 2 x 2 + 2a 2 xc + a 2c 2 + a 2 y 2 = a 4 + 2a 2 xc + x 2c 2 Combine like terms

a x − x c +a y = a −a c
2 2 2 2 2 2 4 2 2
Factor common terms
(a 2 2
)
−c x +a y = a a −c
2 2 2 2
( 2 2
)
Let b 2 = a 2 − c 2 . Since a > c, we know b > 0. Substituting b 2 for a 2 − c 2 leaves
b 2 x 2 + a 2 y 2 = a 2b 2 Divide both sides by a 2b 2
x2 y2
+ =1
a 2 b2

This is the standard equation for an ellipse. We typically swap a and b when the major
axis of the ellipse is vertical.

Equation of an Ellipse Centered at the Origin in Standard Form

The standard form of an equation of an ellipse centered at the origin C(0,0) depends
on whether the major axis is horizontal or vertical. The table below gives the standard
equation, vertices, minor axis endpoints, foci, and graph for each.

Major Axis Horizontal Vertical

2 2 2
Standard x y x y2
+ 2 =1 + =1
Equation a 2
b b2 a 2

Vertices (−a, 0) and (a, 0) (0, −a) and (0, a)

Minor Axis
(0, −b) and (0, b) (−b, 0) and (b, 0)
Endpoints
(−c, 0) and (c, 0) (0, −c) and (0, c)
Foci
where b 2 = a 2 − c 2 where b 2 = a 2 − c 2
y
Graph (0,a)
y
(0,b)
(0,c)
(-a,0) (-c,0) (c,0) (a,0) (-b,0) (b,0)
x x

(0,-c)
(0,-b)
(0,-a)
 Section 9.1 Ellipses 583

Example 1
Put the equation of the ellipse 9 x 2 + y 2 = 9 in standard form. Find the vertices, minor
axis endpoints, length of the major axis, and length of the minor axis. Sketch the graph,
then check using a graphing utility.

The standard equation has a 1 on the right side, so this equation can be put in standard
form by dividing by 9:
x2 y2
+ =1
1 9

Since the y-denominator is greater than the x-denominator, the ellipse has a vertical
x2 y2
major axis. Comparing to the general standard form equation 2 + 2 = 1 , we see the
b a
value of a = 9 = 3 and the value of b = 1 = 1.

The vertices lie on the y-axis at (0,±a) = (0, ±3).

The minor axis endpoints lie on the x-axis at (±b, 0) = (±1, 0).
The length of the major axis is 2(a ) = 2(3) = 6 .
The length of the minor axis is 2(b) = 2(1) = 2 .

To sketch the graph we plot the vertices and the minor axis endpoints. Then we sketch
the ellipse, rounding at the vertices and the minor axis endpoints.

To check on a graphing utility, we must solve the equation for y. Isolating y 2 gives us
(
y 2 = 9 1− x2 )
Taking the square root of both sides we get
y = 3 1 − x 2

Under Y= on your graphing utility enter the two halves of the ellipse as y = 3 1 − x 2
and y = −3 1 − x 2 . Set the window to a comparable scale to the sketch with xmin = -5,
xmax = 5, ymin= -5, and ymax = 5.
584 Chapter 9

Here’s an example output on a TI-84 calculator:

Sometimes we are given the equation. Sometimes we need to find the equation from a
graph or other information.

Example 2
Find the standard form of the equation for an ellipse centered at (0,0) with horizontal
major axis length 28 and minor axis length 16.

Since the center is at (0,0) and the major axis is horizontal, the ellipse equation has the
x2 y2
standard form 2 + 2 = 1 . The major axis has length 2a = 28 or a = 14. The minor
a b
x2 y 2 x2 y2
axis has length 2b = 16 or b = 8. Substituting gives + = 1 or + = 1.
142 82 196 64.

Try it Now
1. Find the standard form of the equation for an ellipse with horizontal major axis length
20 and minor axis length 6.

Example 3
Find the standard form of the equation for the ellipse graphed here.

The center is at (0,0) and the major axis is vertical, so the standard
x2 y2
form of the equation will be 2 + 2 = 1 .
b a

From the graph we can see the vertices are (0,4) and (0,-4), giving
a = 4.
The minor-axis endpoints are (2,0) and (-2,0), giving b = 2.

x2 y2 x2 y2
The equation will be + = 1 or + = 1.
22 42 4 16
 Section 9.1 Ellipses 585

Ellipses Not Centered at the Origin

Not all ellipses are centered at the origin. The graph of such an ellipse is a shift of the
graph centered at the origin, so the standard equation for one centered at (h, k) is slightly
different. We can shift the graph right h units and up k units by replacing x with x – h
and y with y – k, similar to what we did when we learned transformations.

Equation of an Ellipse Centered at (h, k) in Standard Form

The standard form of an equation of an ellipse centered at the point C (h, k ) depends
on whether the major axis is horizontal or vertical. The table below gives the standard
equation, vertices, minor axis endpoints, foci, and graph for each.

Major Axis Horizontal Vertical

Standard (x − h )2 + ( y − k )2 =1
(x − h )2 + ( y − k )2 =1
Equation a2 b2 b2 a2

Vertices ( h ± a, k ) (h, k ± a)

Minor Axis
( h, k ± b ) ( h ± b, k )
Endpoints
( h ± c, k ) (h, k ± c)
Foci
where b2 = a2 – c2 where b2 = a2 – c2
y (h,k+a)
y
(h,k+b)
(h,k+c)
(h-b,k) (h,k) (h+b,k)
Graph (h-a,k) (h-c,k) (h+c,k) (h+a,k)
(h,k)
x
x (h, k-c)
(h,k-b)
(h, k-a)
586 Chapter 9

Example 4
Put the equation of the ellipse x 2 + 2 x + 4 y 2 − 24 y = −33 in standard form. Find the
vertices, minor axis endpoints, length of the major axis, and length of the minor axis.
Sketch the graph.

To rewrite this in standard form, we will need to complete the square, twice.

Looking at the x terms, x 2 + 2 x , we like to have something of the form ( x + n) 2 . Notice

that if we were to expand this, we’d get x 2 + 2nx + n 2 , so in order for the coefficient on
x to match, we’ll need ( x + 1) 2 = x 2 + 2 x + 1 . However, we don’t have a +1 on the left
side of the equation to allow this factoring. To accommodate this, we will add 1 to both
sides of the equation, which then allows us to factor the left side as a perfect square:
x 2 + 2 x + 1 + 4 y 2 − 24 y = −33 + 1
( x + 1) 2 + 4 y 2 − 24 y = −32

Repeating the same approach with the y terms, first we’ll factor out the 4.
4 y 2 − 24 y = 4( y 2 − 6 y)

( ) ( )
Now we want to be able to write 4 y 2 − 6 y as 4( y + n) 2 = 4 y 2 + 2ny + n 2 .
For the coefficient of y to match, n will have to -3, giving
( )
4( y − 3) 2 = 4 y 2 − 6 y + 9 = 4 y 2 − 24 y + 36 .

To allow this factoring, we can add 36 to both sides of the equation.

( x + 1) 2 + 4 y 2 − 24 y + 36 = −32 + 36
( )
( x + 1) 2 + 4 y 2 − 6 y + 9 = 4
( x + 1) 2 + 4( y − 3) = 4
2

Dividing by 4 gives the standard form of the equation for the ellipse
(x + 1)2 + ( y − 3)2 = 1
4 1

Since the x-denominator is greater than the y-denominator, the ellipse has a horizontal
( x − h) ( y − k )
2 2

major axis. From the general standard equation + = 1 we see the value
a2 b2
of a = 4 = 2 and the value of b = 1 = 1.

The center is at (h, k) = (-1, 3).

The vertices are at (h±a, k) or (-3, 3) and (1,3).
The minor axis endpoints are at (h, k±b) or (-1, 2) and (-1,4).
 Section 9.1 Ellipses 587

The length of the major axis is 2(a) = 2(2) = 4 .

The length of the minor axis is 2(b) = 2(1) = 2 .

To sketch the graph we plot the vertices and the minor axis endpoints. Then we sketch
the ellipse, rounding at the vertices and the minor axis endpoints.

Example 5
Find the standard form of the equation for an ellipse centered at (-2,1), a vertex at (-2,4)
and passing through the point (0,1).

The center at (-2,1) and vertex at (-2,4) means the major axis is vertical since the x-

values are the same. The ellipse equation has the standard form
(x − h )2 + ( y − k )2 = 1 .
b2 a2

The value of a = 4-1=3. Substituting a = 3, h = -2, and k = 1 gives

(x + 2)2 + ( y − 1)2 = 1 . Substituting for x and y using the point (0,1) gives
b2 32
(0 + 2)2 + (1 − 1)2 = 1 .
b2 32
Solving for b gives b=2.

The equation of the ellipse in standard form is

( x + 2) ( y − 1)
2
+
2
= 1 or
22 32
(x + 2)2 + ( y − 1)2 = 1 .
4 9

Try it Now
2. Find the center, vertices, minor axis endpoints, length of the major axis, and length of

the minor axis for the ellipse ( x − 4) +

2 ( y + 2)2 = 1 .
4
588 Chapter 9

Bridges with Semielliptical Arches

Arches have been used to build bridges for centuries, like

in the Skerton Bridge in England which uses five
semielliptical arches for support3. Semielliptical arches
can have engineering benefits such as allowing for longer
spans between supports.

Example 6
A bridge over a river is supported by a single semielliptical arch. The river is 50 feet
wide. At the center, the arch rises 20 feet above the river. The roadway is 4 feet above
the center of the arch. What is the vertical distance between the roadway and the arch
15 feet from the center?

Put the center of the ellipse at (0,0) and make the span of the river the major axis.

4ft

20ft

x
50ft

x2 y2
Since the major axis is horizontal, the equation has the form + = 1.
a 2 b2
1 x2 y2
The value of a = (50 ) = 25 and the value of b = 20, giving + = 1.
2 252 202
15 2 y2 225
Substituting x = 15 gives 2
+ 2
= 1 . Solving for y, y = 20 1 − = 16 .
25 20 625

The roadway is 20 + 4 = 24 feet above the river. The vertical distance between the
roadway and the arch 15 feet from the center is 24 − 16 = 8 feet.

3
Maxine Armstrong
(https://commons.wikimedia.org/wiki/File:Skerton_Bridge,_Lancaster,_England.JPG), “Skerton Bridge,
Lancaster, England”, CC BY-SA
 Section 9.1 Ellipses 589

Ellipse Foci

The location of the foci can play a key role in ellipse application problems. Standing on a
focus in a whispering gallery allows you to hear someone whispering at the other focus.
To find the foci, we need to find the length from the center to the foci, c, using the
equation b 2 = a 2 − c 2 . It looks similar to, but is not the same as, the Pythagorean
Theorem.

Example 7
The National Statuary Hall whispering chamber is an elliptical room 46 feet wide and
96 feet long. To hear each other whispering, two people need to stand at the foci of the
ellipse. Where should they stand?

We could represent the hall with a horizontal ellipse centered at the origin. The major
1
axis length would be 96 feet, so a = (96 ) = 48 , and the minor axis length would be 46
2
1
feet, so b = ( 46 ) = 23 . To find the foci, we can use the equation b 2 = a 2 − c 2 .
2
23 = 48 − c 2
2 2

c 2 = 48 2 − 23 2
c = 1775  42 ft.

To hear each other whisper, two people would need to stand 2(42) = 84 feet apart along
the major axis, each about 48 – 42 = 6 feet from the wall.

Example 8

Find the foci of the ellipse

(x − 2)2 + ( y + 3)2 = 1.
4 29

The ellipse is vertical with an equation of the form

(x − h )2 + ( y − k )2 =1.
b2 a2
The center is at (h, k) = (2, −3). The foci are at (h, k ± c).

To find length c we use b 2 = a 2 − c 2 .

Substituting gives 4 = 29 − c 2 or c = 25 = 5 .

The ellipse has foci (2, −3 ± 5), or (2, −8) and (2, 2).
590 Chapter 9

Example 9
Find the standard form of the equation for an ellipse with foci (-1,4) and (3,4) and major
axis length 10.

Since the foci differ in the x -coordinates, the ellipse is horizontal with an equation of

the form
(x − h )2 + (h − k )2 = 1 .
a2 b2
 x + x y + y2   (− 1) + 3 4 + 4 
The center is at the midpoint of the foci  1 2 , 1 = ,  = (1, 4 ) .
 2 2   2 2 
1
The value of a is half the major axis length: a = (10 ) = 5 .
2
1 1
The value of c is half the distance between the foci: c = (3 − (−1)) = (4) = 2 .
2 2
To find length b we use b = a − c . Substituting a and c gives b = 52 − 2 2 = 21.
2 2 2 2

The equation of the ellipse in standard form is

(x − 1)2 + ( y − 4)2 = 1 or
52 21
(x − 1)2 + ( y − 4)2 =1.
25 21

Try it Now
3. Find the standard form of the equation for an ellipse with focus (2,4), vertex (2,6),
and center (2,1).

Planetary Orbits

It was long thought that planetary orbits around the

sun were circular. Around 1600, Johannes Kepler
discovered they were actually elliptical4. His first law
of planetary motion says that planets travel around the
sun in an elliptical orbit with the sun as one of the foci.
The length of the major axis can be found by measuring the planet’s aphelion, its greatest
distance from the sun, and perihelion, its shortest distance from the sun, and summing
them together.

4
Technically, they’re approximately elliptical. The orbits of the planets are not exactly elliptical because
of interactions with each other and other celestial bodies.
 Section 9.1 Ellipses 591

Example 10
Mercury’s aphelion is 35.98 million miles and its perihelion is 28.58 million miles.
Write an equation for Mercury’s orbit.

Let the center of the ellipse be (0,0) and its major axis be horizontal so the equation will
x2 y2
have form 2 + 2 = 1 .
a b

The length of the major axis is 2a = 35.98 + 28.58 = 64.56 giving a = 32.28 and
a 2 = 1041 .9984 .

Since the perihelion is the distance from the focus to one vertex, we can find the
distance between the foci by subtracting twice the perihelion from the major axis
length: 2c = 64.56 − 2(28.58) = 7.4 giving c = 3.7 .
Substitution of a and c into b 2 = a 2 − c 2 yields b 2 = 32 .28 2 − 3.7 2 = 1028 .3084 .

x2 y2
The equation is + =1.
1041 .9984 1028 .3084

Important Topics of This Section

Ellipse Definition
Ellipse Equations in Standard Form
Ellipse Foci
Applications of Ellipses

Try it Now Answers

x2 y2
1. 2a = 20, so a =10. 2b = 6, so b = 3. + =1
100 9

2. Center (4, -2). Vertical ellipse with a = 2, b = 1.

Vertices at (4, -2±2) = (4,0) and (4,-4),
minor axis endpoints at (4±1, -2) = (3,-2) and (5,-2),
major axis length 4, minor axis length 2

3. Vertex, center, and focus have the same x-value, so it’s a vertical ellipse.
Using the vertex and center, a = 6 – 1 = 5
Using the center and focus, c = 4 – 1 = 3
592 Chapter 9

b 2 = 52 − 32 . b = 4.
(x − 2)2 + ( y − 1)2 = 1
16 25
 Section 9.1 Ellipses 593

Section 9.1 Exercises

In problems 1–4, match each graph with one of the equations A–D.
x2 y2 x2 y2 x2 y2
A. + =1 B. + =1 C. + y2 = 1 D. x 2 + =1
4 9 9 4 9 9

1. 2. 3. 4.

In problems 5–14, find the vertices, the minor axis endpoints, length of the major axis,
and length of the minor axis. Sketch the graph. Check using a graphing utility.
x2 y2 x2 y2 x2 y2
5. + =1 6. + =1 7. + y =1
2
8. x +2
=1
4 25 16 4 4 25

9. x 2 + 25 y 2 = 25 10. 16 x 2 + y 2 = 16 11. 16 x 2 + 9 y 2 = 144

12. 16 x 2 + 25 y 2 = 400 13. 9 x 2 + y 2 = 18 14. x 2 + 4 y 2 = 12

In problems 15–16, write an equation for the graph.

15. 16.

In problems 17–20, find the standard form of the equation for an ellipse satisfying the
given conditions.
17. Center (0,0), horizontal major axis length 64, minor axis length 14

18. Center (0,0), vertical major axis length 36, minor axis length 18

19. Center (0,0), vertex (0,3), b = 2

20. Center (0,0), vertex (4,0), b = 3

594 Chapter 9

In problems 21–28, match each graph to equations A-H.

A.
(x − 2)2 + ( y − 1) 2 = 1 E.
(x + 2)2 + ( y + 1) 2 = 1
4 9 4 9

B.
(x − 2) + ( y − 1) = 1
2 2
F.
(x + 2) + ( y + 1) 2 = 1
2

4 16 4 16

C.
(x − 2) + ( y − 1) = 1
2 2
G.
(x + 2) + ( y + 1) 2 = 1
2

16 4 16 4

D.
(x − 2) + ( y − 1) = 1
2 2
H.
(x + 2) + ( y + 1) 2 = 1
2

9 4 9 4

21. 22. 23. 24.

25. 26. 27. 28.

In problems 29–38, find the vertices, the minor axis endpoints, length of the major axis,
and length of the minor axis. Sketch the graph. Check using a graphing utility.
( x − 1) 2 ( y + 2) 2 ( x + 5) 2 ( y − 3) 2
29. + =1 30. + =1
25 4 16 36

( y − 3) 2 ( x − 1) 2
31. ( x + 2) 2 + =1 32. + ( y − 6) 2 = 1
25 25

33. 4 x 2 + 8x + 4 + y 2 = 16 34. x 2 + 4 y 2 + 16 y + 16 = 36

35. x 2 + 2 x + 4 y 2 + 16 y = −1 36. 4 x 2 + 16 x + y 2 − 8 y = 4

37. 9 x 2 − 36 x + 4 y 2 + 8 y = 104 38. 4 x 2 + 8x + 9 y 2 + 36 y = −4

 Section 9.1 Ellipses 595

In problems 39–40, write an equation for the graph.

39. 40.

In problems 41–42, find the standard form of the equation for an ellipse satisfying the
given conditions.
41. Center (-4,3), vertex(-4,8), point on the graph (0,3)

42. Center (1,-2), vertex(-5,-2), point on the graph (1,0)

43. Window A window in the shape of a semiellipse is 12 feet wide and 4 feet high.
What is the height of the window above the base 5 feet from the center ?

44. Window A window in the shape of a semiellipse is 16 feet wide and 7 feet high.
What is the height of the window above the base 4 feet from the center?

45. Bridge A bridge over a river is supported by a semielliptical arch. The river is 150
feet wide. At the center, the arch rises 60 feet above the river. The roadway is 5 feet
above the center of the arch. What is the vertical distance between the roadway and
the arch 45 feet from the center?

46. Bridge A bridge over a river is supported by a semielliptical arch. The river is 1250
feet wide. At the center, the arch rises 175 feet above the river. The roadway is 3
feet above the center of the arch. What is the vertical distance between the roadway
and the arch 600 feet from the center?

47. Racetrack An elliptical racetrack is 100 feet long and 90 feet wide. What is the
width of the racetrack 20 feet from a vertex on the major axis?

48. Racetrack An elliptical racetrack is 250 feet long and 150 feet wide. What is the
width of the racetrack 25 feet from a vertex on the major axis?
596 Chapter 9

In problems 49-52, find the foci.

x2 y2 x2 y2
49. + =1 50. + =1
19 3 2 38

( y − 1) 2 ( x − 3) 2
51. ( x + 6) 2 + + =1 52. + ( y + 5) 2 = 1
26 10

In problems 53-72, find the standard form of the equation for an ellipse satisfying the
given conditions.

53. Major axis vertices (±3,0), c=2 54. Major axis vertices (0,±7), c=4

55. Foci (0,±5) and major axis length 12

56. Foci (±3,0) and major axis length 8

57. Foci (±5,0), vertices (±7,0) 58. Foci (0,±2), vertices (0,±3)

59. Foci (0,±4) and x-intercepts (±2,0)

60. Foci (±3,0) and y-intercepts (0,±1)

61. Center (0,0), major axis length 8, foci on x-axis, passes through point 2, 6 ( )
62. Center (0,0), major axis length 12, foci on y-axis, passes through point ( 10 ,4)
63. Center (-2,1), vertex (-2,5), focus (-2,3)

64. Center (-1,-3), vertex (-7,-3), focus (-4,-3)

65. Foci (8,2) and (-2,2), major axis length 12

66. Foci (-1,5) and (-1,-3), major axis length 14

67. Vertices (3,4) and (3,-6), c= 2

68. Vertices (2,2) and (-4,2), c= 2

69. Center (1,3), focus (0,3), passes through point (1,5)

70. Center (-1,-2), focus (1,-2), passes through point (2,-2)

71. Focus (-15,-1), vertices (-19,-1) and (15,-1)

72. Focus (-3,2), vertices (-3,4) and (-3,-8)

 Section 9.1 Ellipses 597

73. Whispering Gallery If an elliptical whispering gallery is 80 feet long and 25 feet
wide, how far from the center of room should someone stand on the major axis of
the ellipse to experience the whispering effect? Round to two decimal places.

74. Billiards Some billiards tables are elliptical and have the foci marked on the table. If
such a one is 8 feet long and 6 feet wide, how far are the foci from the center of the
ellipse? Round to two decimal places.

75. Planetary Orbits The orbits of planets around the sun are approximately elliptical
with the sun as a focus. The aphelion is a planet’s greatest distance from the sun and
the perihelion is its shortest. The length of the major axis is the sum of the aphelion
and the perihelion. Earth’s aphelion is 94.51 million miles and its perihelion is
91.40 million miles. Write an equation for Earth’s orbit.

76. Satellite Orbits The orbit of a satellite around Earth is elliptical with Earth’s center
as a focus. The satellite’s maximum height above the Earth is 170 miles and its
minimum height above the Earth is 90 miles. Write an equation for the satellite’s
orbit. Assume Earth is spherical and has a radius of 3960 miles.

c
77. Eccentricity e of an ellipse is the ratio where c is the distance of a focus from the
a
center and a is the distance of a vertex from the center. Write an equation for an
ellipse with eccentricity 0.8 and foci at (-4,0) and (4,0).

78. Confocal ellipses have the same foci. Show that, for k > 0, all ellipses of the form
x2 y2
+ = 1 are confocal.
6+k k

79. The latus rectum of an ellipse is a line segment with endpoints on the ellipse that
2b 2
passes through a focus and is perpendicular to the major axis. Show that is the
a
x2 y2
length of the latus rectum of 2 + 2 = 1 where a > b.
a b