Introduction to Ellipses
Most ellipses look like flattened circles. Usually one diameter is longer than the other. In Figure 12.11, the horizontal diameter is longer than the vertical diameter. In Figure 12.12 the vertical diameter is longer than the horizontal diameter. The longer diameter is the major axis , and the shorter diameter is the minor axis . An ellipse has seven important points—the center, two endpoints of the major axis (the vertices), two endpoints of the minor axis, and two points along the major axis called the foci (plural for focus ). When the equation of an ellipse is in the form
we can find these points without much trouble.
Fig. 12.11
Fig. 12.12
Sketching Ellipses Graphs
If all we want to do is to sketch the graph, all we really need to do is to plot the endpoints of the diameters and draw a rounded curve connecting them. For example, if we want to sketch the graph of 
, a = 3, b = 2, h = −1, and k = 1. According to Figure 12.12, the diameters have coordinates (−1 −2, 1) = (−3, 1), (−1 + 2, 1) = (1, 1), (−1, 1 + 3) = (−1, 4), and (−1, 1 − 3) = (−1, −2). (See Figure 12.13.)
Fig. 12.13
DEFINITION: An ellipse is the set of all points whose distances to two fixed points (the foci) is constant.
For example, the foci for 
. If we take any point on this ellipse and calculate its distance to (−4, 0) and to (4, 0) and add these numbers, the sum will be 10. Two points on this ellipse are (0, 3) and 
.
Distance from (0, 3) to (−4, 0) + Distance from (0, 3) to (4, 0)
Distance from 
to (−4, 0) + Distance from to (4, 0)
Finding the Center, Foci, and Vertices of Ellipses Examples
In the next set of problems, we will be given an equation for an ellipse. From the equation, we can find h, k, a , and b . With these numbers and the information in Figures 12.11 or 12.12 we can find the center, foci, and vertices.
Fig. 12.11
Fig. 12.12
Examples
Find the center, foci, and vertices for the ellipse.
-
From the equation, we see that h = 3, k = −5, a 2 and b 2 are 4 2 and 5 2 , but which is a and which is b ? a needs to be the larger number, so a = 5 and b = 4. This makes 
. We need to use the information in Figure 12.12 because the larger denominator is under ( y − k ) 2 .
Center: ( h , k ) = (3, −5)
Foci: ( h , k − c ) = (3, −5 − 3) = (3, −8) and ( h , k + c ) = (3, −5 + 3) = (3, -2)
Vertices: ( h , k − a ) = (3, −5−5) = (3, −10) and ( h , k + a ) = (3, −5 + 5) = (3, 0)
-
To make it easier to find h, k, a , and b , we will rewrite the equation.
Now we can see that h = 0, k = 2, a = 4, b = 1, 
. Because the larger denominator is under ( x − 0) 2 , we need to use the information in Figure 12.11 .
Center: ( h , k ) = (0, 2)
Foci: 
Vertices: ( h − a , k ) = (0 − 4, 2) = (−4, 2) and ( h + a , k ) = (0 + 4, 2) = (4, 2)
Matching the Equation of an Ellipse to its Graph Examples
Now that we can find this important information from an equation of an ellipse, we are ready to match graphs of ellipses to their equations.
Examples
Match the equations with the graphs in Figures 12.14-12.16.
Fig. 12.14
Fig. 12.15
Fig. 12.16
-
The larger denominator is under y 2 , so we need to use the information in Figure 12.12 . Because a = 3, we need to look for a graph with vertices (0, 3) and (0, −3). This graph is in Figure 12.15.
-
Fig. 12.15
-
The larger denominator is under x 2 , so we need to use the information in Figure 12.11 . Because a = 3, the vertices are (−3, 0) and (3, 0). This graph is in Figure 12.16.
-
Fig. 12.16
-
The larger denominator is under y 2 , so we need to use the information in Figure 12.12 . Because a = 2, the vertices are (0, 2) and (0, −2). This graph is in Figure 12.14.
-
Fig. 12.14
Finding the Equation of Ellipses
With as little as three points, we can find an equation of an ellipse. Using the formulas in Figures 12.11 and 12.12 and some algebra, we can find h, k, a , and b .
Fig. 12.11
Fig. 12.12
Examples
Find an equation of the ellipse.
- The vertices are (−4, 2) and (6, 2), and (1, 5) is a point on the graph. The y -coordinates are the same, so the major axis (the larger diameter) is horizontal, which means we need to use the information in Figure 12.11. The vertices are ( h − a, k ) and ( h + a, k ). This means that h − a = −4 and h + a = 6, and k = 2.
So far we know that
Because (1, 5) is on the graph, 
. Solving this equation for b , we find that b = 3. The equation is
.
- The foci are (−4, −10) and (−4, 14) and (−4, 15) is a vertex.
-
The x -values of foci are the same, so the major axis is vertical. This tells us that we need to use the information in Figure 12.12 .
( h , k − c ) = (−4, −10) and ( h , k + c ) = (−4, 14), so h = −4, k − c = −10 and k + c = 14.
Because (−4, 15) is a vertex, k + a = 15, so 2 + a = 15 and a = 13. All we need to finish is to find b . Let a = 13 and c = 12 in 
: 
. Solving this for b , we have b = 5. The equation is
Finding the Eccentricity of Ellipses
The eccentricity of an ellipse is a number that measures how flat it is. The formula is 
. This number ranges between 0 and 1. The closer to 1 the eccentricity of an ellipse is, the flatter it is. If 
, then the ellipse is a circle. In a circle, the center and foci are all the same point, and a and b are the same number. For example, 
is a circle with center (5, 4) and radius 
. Usually we see equations of circles in the form ( x − h ) 2 + ( y − k ) 2 = r 2 .
Examples
Find the ellipse’s eccentricity.
a = 13, b = 12, 
, 
This ellipse is more rounded than the first because e is closer to 0.
Ellipses Practice Problems
Practice
-
Identify the center, foci, vertices, and eccentricity for
-
Identify the center, foci, vertices, and eccentricity for
-
Identify the center and radius for the circle
For Problems 4-7, match the equation with the graph in Figures 12.17-12.20.
Fig. 12.17
Fig. 12.18
Fig. 12.19
Fig. 12.20
Solutions
-
h = 0, k = 10, a = 13, b = 5, 
Center: (0, 10)
Foci: ( h − c , k ) = (0−12, 10) = (−12, 10) and ( h + c , k ) = (0 + 12, 10) (12, 10)
Vertices: ( h − a , k ) = (0 − 13, 10) = (−13, 10) and ( h + a , k ) (0 + 13, 10) = (13, 10)
Eccentricity: 
-
h = −9, k = −2, a = 29, b = 20, 
Center: (−9, −2)
Foci: ( h , k − c ) = (−9, −2−21) = (−9, −23) and ( h , k + c ) = (−9, −2 + 21) = (−9, 19)
Vertices: ( h , k − a ) = (−9, −2 − 29) = (−9, −31) and ( h , k + a ) = (−9, −2 + 29) = (−9, 27)
Eccentricity: 
-
The center is (−6, 1), and the radius is 7.
-
Figure 12.19
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Figure 12.20
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Figure 12.18
-
Figure 12.17
Practice problems for this concept can be found at: Conic Sections Practice Test.
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