Ellipses
By Thomas Martin and
Sarah Baldree
A conic section is a group of points that moves in the plane of a fixed point (focus) and a fixed
line (directrix). An ellipse is a conic section that looks like an elongated circle, or oval.
Where do circles, ellipses,
hyperbolas and parabolas like
to hang out in the summer?
Coney Island!
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
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There are two types of ellipses: horizontal and vertical. Ellipses are determined as
horizontal or vertical by the direction of the major axis. So, if the major axis, or the
greater distance across the ellipse, is horizontal, the ellipse is called a horizontal ellipse.
However, if the major axis is vertical, the ellipse is considered a vertical ellipse. In the
picture above, the ellipse on the left side, ellipse a, is a horizontal ellipse, while the
ellipse on the right side, ellipse b, is a vertical ellipse.
The standard form of the equation for an ellipse is determined by whether the ellipse is
horizontal or vertical.
-The standard form equation for a horizontal ellipse is
(𝑥 − ℎ)2 (𝑦 − 𝑘)2
+
=1
𝑎2
𝑏2
-The standard form equation for a vertical ellipse is
(𝑥 − ℎ)2 (𝑦 − 𝑘)2
+
=1
𝑏2
𝑎2
When you are only given the equation of an ellipse, it can be difficult to tell whether the
ellipse is vertical or horizontal. However, the 𝑎 value is always larger than the 𝑏 value,
which means that when the denominator of the x value is larger than the denominator of
the y value, the ellipse is horizontal. When the denominator of the x value is smaller
than the denominator of the y value, the ellipse is vertical.
In the standard form equations, the variables are defined as follows:

 𝑎: One half of the length of the major axis, or the distance from the center to either
vertex.
 𝑏: One half of the length of the minor axis, or the distance from the center to either
co-vertex.
 ℎ: The x component of the center of the ellipse.
 𝑘: The y component of the center of the ellipse.
Ellipses have six main components:
 Major axis: The chord that runs through the foci and intersects the ellipse at two
points (The vertices). The length of the major axis is equal to 2𝑎. The major axis is
the longer of the two axes.
 Minor axis: The chord that runs perpendicular to the major axis and intersects the
ellipse at two points (The co-vertices). The length of the minor axis is equal to 2𝑏.
The minor axis is the shorter of the two axes.
Here is a horizontal ellipse
and its axes.
Major
Minor

 Foci: The two points that dictate the eccentricity of the ellipse. The foci are always
located on the major axis. To find the foci use the equation 𝑐 2 = 𝑎2 − 𝑏 2 . When you
find 𝑐, you can find the foci. For a vertical ellipse, the foci are at the points
(ℎ, 𝑘 ± 𝑐), and for a horizontal ellipse the foci are at the points(ℎ ± 𝑐, 𝑘).
 Vertices: Two points that lie on the major axis, or the line that runs through the
focus. They are both 𝑎 units away from the center of the ellipse.
 Co-vertices: Two points that lie on the minor axis where it intersects the ellipse.
They are both 𝑏 units away from the center of the ellipse.
 Center: The center is the point at which the major and minor axis meet. It is also at
the center of the ellipse, which is why it is called the center (duh). It is at the point
(ℎ, 𝑘).
An ellipse’s shape is defined by its eccentricity. Eccentricity is a number between 0 and
1. The more circular the ellipse is, the closer the eccentricity is to 0. The more elongated
the ellipse is, the closer the eccentricity is to 1.


𝑐
Eccentricity, or 𝑒, is found using the equation 𝑒 = (𝑎). If you don’t know what 𝑐 is, you
can use the equation 𝑐 2 = 𝑎2 − 𝑏 2 .
If you need further explanation, you can look on pages 353 and 354 in the math
textbook, or go to Khan Academy, where they offer multiple videos and tutorials on
ellipses.
Example problems:
1. A furniture manufacturer wishes to make elliptical tops for coffee tables 52 in. long and 26 in.
wide. A pattern is to be cut from plywood so that the outline of the tabletops can easily be
marked on the tabletop’s surface. Find the equation of the ellipse and its eccentricity.
-The first step in writing the equation is deciding whether it’s a horizontal or vertical ellipse so
we know which version of the equation to use. It says that it’s 52 in. long but only 26 in. wide, so
you know it’s a horizontal ellipse and you use the equation:
(𝑥 − ℎ)2 (𝑦 − 𝑘)2
+
=1
𝑎2
𝑏2
-To make things easy, we can assume that the center of the ellipse would be at the origin or (0,0).
In the equation we know (ℎ, 𝑘) is the center, so we just plug in (0,0) for those variables. Now the
equation looks like:
(𝑥 − 0)2 (𝑦 − 0)2
+
=1
𝑎2
𝑏2
Or more simply:
𝑥2 𝑦2
+
=1
𝑎2 𝑏 2
-Next, you should find the values of 𝑎 and 𝑏. We know that 2𝑎 = the length of the major axis.
We already know the length of the major axis (it’s 52 inches) so we can use that to find 𝑎, like
so:
2𝑎 = 52
Then divide both sides by 2 and you find that:
𝑎 = 26
-You can then apply the same principles to find 𝑏.
2𝑏 = 26
𝑏 = 13
Now that you know 𝑎 and 𝑏, you can plug them into the equation we already have.
𝑥2
𝑦2
+
=1
262 132
And then simplify it further to find the final equation.
𝑥2
𝑦2
+
=1
676 169
-For the final step of this problem, we have to find the eccentricity. We know that to find the
𝑐
eccentricity, you use the equation 𝑒 = (𝑎). First, you need to find the value of 𝑐, which can be
done by manipulating the equation 𝑐 2 = 𝑎2 − 𝑏 2 . Since we already found 𝑎 and 𝑏, all you have
to do is plug them into the equation to find 𝑐, like so:
𝑐 2 = 262 − 132
Then simplify that equation:
𝑐 2 = 676 − 169
𝑐 2 = 507
𝑐 = √507
𝑐
Then use the original eccentricity equation, 𝑒 = (𝑎).
𝑒=
√507
26
𝑒 = .86602
Looking at that number, we know that since 𝑒 is closer to 1 than to 0 that the ellipse is more
elongated than circular, and now we’ve finished the problem.
2. Write the equation of an ellipse with vertices of (0,-8) and (0,8) and co-vertices of (-5,0) and
(5,0).
To solve this problem, you must first decide whether the ellipse is horizontal or vertical. The
vertices and co-vertices have been provided for this problem, making it rather easy. We know
that the vertices mark the endpoints of the major axis, so by plotting them we could see if the
axis was vertical or horizontal and therefore know what equation to use. The vertices are oriented
vertically on the y axis, which would make the major axis vertical, which means that the
equation for a vertical ellipse would be used:
(𝑥 − ℎ)2 (𝑦 − 𝑘)2
+
=1
𝑏2
𝑎2
Next we would find the values of 𝑎 and 𝑏. To do so, we have to find the distance between the
vertices, which would be the length of the major axis. We know that the length of the major axis
is equal to 2𝑎, so by dividing the length of the major axis by 2, we could find 𝑎. Since the 2
vertices are both 8 units away from a central point (the center), which happens to be the origin in
this problem, the major axis would be 16 units long and 𝑎 would be 8.
2𝑎 = 16
𝑎=8
We also know that the length of the minor axis is equal to 2𝑏, so by dividing the length of the
minor axis by 2, we could find 𝑏. Since the 2 co-vertices are both 5 units away from a central
point (the center), the minor axis would be 10 units long and 𝑏 would be 5.
2𝑏 = 10
𝑏=5
Next, we need to find the center. We know that the center would be located midway between the
two vertices, so one could use the midpoint formula. When you use the midpoint formula, you
add the components of two coordinate points and divide them by two, effectively finding the
point between them.
(ℎ, 𝑘) =
(0, −8) + (0,8)
2
(ℎ, 𝑘) = (0,0)
So now that we have 𝑎, 𝑏, and the center, we can plug these values into the equation of a vertical
ellipse, like so:
(𝑥 − 0)2 (𝑦 − 0)2
+
=1
52
82
Then, simplify:
𝑥2 𝑦2
+
=1
25 64
3. Find the eccentricity of the following equation:
𝑐
We know that to find eccentricity, you use the equation 𝑒 = (𝑎). The value of 𝑎 is easiest to find,
so we can start with that. If you may recall, the value for 𝑎 is found in the denominator of one of
the two fractions, and it is always the larger of the two denominators. However, it is very
important to remember to take the square root of the larger number, or we’ll be left with 𝑎2
rather than just 𝑎. So by looking at the equation, we see that 36 is the larger denominator, and
√36 is equal to 6. Now we’ve filled in one of the variables, and the equation now looks like this:
𝑐
𝑒 = (6).
Next we have to find 𝑐. To do so, you use the equation that relates 𝑎, 𝑏, and 𝑐. That equation
is 𝑐 2 = 𝑎2 − 𝑏 2 . Since in the ellipse equation 𝑎 and 𝑏 are already squared, all we have to do is
take the two denominators and subtract the smaller from the larger, like so:
𝑐 2 = 36 − 16
𝑐 2 = 20
𝑐 = √20
Then, we would plug that into the equation we’re forming and we would have this:
𝑒=(
√20
)
6
And simplify:
𝑒 = 0.75
The equation is completely simplified, and we are done with this problem, and have learned that
this ellipse is more oblong than round, since the value of 𝑒 is closer to 1 than 0.
Now you’re ready to try your own problems!!!
Practice Problem #1:
Find the vertices and co-vertices of this equation:
Practice Problem #2
Write the equation for this ellipse:
Practice Problem #3
Find the eccentricity of this equation:
When you’ve finished these problems on your own, you can compare your answers to those
provided on the next page.
Solutions
Practice Problem #1:
Find the vertices and co-vertices of this equation:
By examining the equation, we see that the denominator of the y value is larger than the
denominator of the x value. This means that this is a vertical ellipse. Next, we can find the
ellipse’s center. Since there are no values for h or k, they are assumed to be zero. So, the center is
at the origin, (0,0). Next, we need to find a and b.
Using the formula:
(𝑥 − 0)2 (𝑦 − 0)2
+
=1
36
100
We can find the value of a by taking the square root of 100, which is 10.
We can find the value of b by taking the square root of 36, which is 6.
So now that we have that 𝑎 = 10 and 𝑏 = 6, we can find the vertices and co-vertices.
For a vertical ellipse, the vertices are represented by the points (ℎ, 𝑘 ± 𝑎). If we plug in our
values, we find that the vertices are (0,0 ± 10), or (0,10) and (0,-10).
For a vertical ellipse, the co-vertices are represented by the points (ℎ ± 𝑏, 𝑘). If we plug in our
values, we find that the co-vertices are (0 ± 6,0), or (6,0) and (-6,0).
Practice Problem #2
Write the equation for this ellipse:
This is a really easy problem to tackle. By simply looking at the graph of the ellipse, we can see
that it is a vertical ellipse, since the longer axis, or major axis, is oriented vertically. Therefore,
we know to use the vertical ellipse equation:
(𝑥 − ℎ)2 (𝑦 − 𝑘)2
+
=1
𝑏2
𝑎2
First, we would find the center and plug it in. The ellipse is shown to have a center, labelled C, at
the point (3,-6). In the ellipse equation, we would plug this is for ℎ and 𝑘. We know that ℎ
corresponds to the x-value, and 𝑘 corresponds to the y-value, so it’s as simple as plugging it in.
Our equation looks like this after we plug it in:
(𝑥 − 3)2 (𝑦 − (−6))2
+
=1
𝑏2
𝑎2
And is simplified as:
(𝑥 − 3)2 (𝑦 + 6)2
+
=1
𝑏2
𝑎2
The next step would be to find the values of 𝑎 and 𝑏, or the distance between one of the vertices
and the center, and the distance between one of the co-vertices and the center. To find such, you
can just subtract the corresponding parts of the coordinates. First we’ll find 𝑎. By looking at the
graph, we see that one of the vertices is located at (3,2), and that the center is at (3,-6). By
subtracting these two we will find 𝑎:
(3,2) − (3, −6) = (0,8)
Therefore,
𝑎=8
Then we can apply the same principle to b. We see that one of the co-vertices is located at (7,-6),
and that the center is at (3,-6). By subtracting the two, we will find 𝑏:
(7, −6) − (3, −6) = (4,0)
Therefore,
𝑏=4
Now all we have to do to finish is plug the values into the equation and simplify, like this:
(𝑥 − 3)2 (𝑦 + 6)2
+
=1
42
82
(𝑥 − 3)2 (𝑦 + 6)2
+
=1
16
64
That is your final answer, and the problem is solved!
Practice Problem #3:
Find the eccentricity of this equation and sketch its graph:
The standard form of an ellipse with a vertical major axis and the center at the origin is:
Comparing the equation with the standard form:
a2 = 32 and b2 = 22.
Taking the square root on both sides:
a = 3 and b = 2
The end points of the major axis of an ellipse (0, a) and (0, –a) are the vertices.
So, if you plug in 3 for 𝑎, you see that the points (0, 3) and (0, –3) are the vertices of the given ellipse.
The end points of the minor axis of an ellipse (b, 0) and (–b, 0) are the co–vertices.
So, if you plug in 2 for 𝑏, you see that the points (2, 0) and (–2, 0) are the co–vertices of the given ellipse.
Using the vertices and co–vertices, the ellipse can be plotted as shown in the figure:
Now that we have the values of a and b, it is a relatively easy matter to find the ellipse’s eccentricity. We
can do this by using the equation 𝑐 2 = 𝑎2 − 𝑏 2 to find c, then plugging it into the original
𝑐
eccentricity equation, , 𝑒 = (𝑎), like so:
𝑐 2 = 32 − 22
𝑐 = √5
𝑒=
√5
3
𝑒 = 0.7453559925
So, the eccentricity of the ellipse is 0.7453559925, which means that it is more elongated than circular,
which is supported by the graph.