 # Ellipse – Properties, Components, and Graph

Ellipses exhibit oval-shaped conic sections. They are actually used to model planets’ orbital motion and have extensive optics, astronomy, and architecture applications.

Ellipses are conic sections that are formed by using an inclined plane to cut through a cone. These sections are oval in shape can contain two foci and vertices.

We’ve discussed ellipses in our conics section, so you can check the article out to have a quick refresher on how ellipses are obtained and are different from the other conic sections.

• How they were formed and the different types of ellipses.
• The general forms for standard and translated ellipses.
• Know how to determine the foci, vertices, and centers of ellipses given their graphs and equations.

Let’s go ahead and dive right into a mental exercise – let’s understand how we can illustrate ellipses’ geometric representation.

## What is an ellipse?

One of the earliest definitions we learned about ellipses is how it results from an inclined plane cutting through a cone (or a double cone, for our example). As can be seen from the image shown above, the resulting section is an oval-shaped conic section or, as we know it, an ellipse. Now that we understand how the shape of an ellipse is formed, it’s time we understand why we have two foci and vertices for this section.

Here’s an exercise we can try before we dive right into a more formal definition of an ellipse. If you have the materials, try actually to perform this exercise.

• Try placing two pins to represent two fixed points (we call each point the ellipse’s focus).
• Let’s say we have a string with a length longer than the distance between the two points. Imagine pinning the ends of that string using the two pins.
• Now, hold a pen or a pencil taut inside the string and trace the curve formed by the pen’s motion.

The curve’s full path will represent the ellipse’s geometric definition. This can help us understand how we need two foci to guide us when graphing ellipse and why ellipses’ major and minor axes have different lengths.

We’ll see also why the sum of the distances between the two foci from any point on the curve will be constant.

Ellipse definition

From the previous exercise, we know that the string’s length will remain the same regardless of the pen’s location. We can extend this to define ellipses formally.

An ellipse is a conic section that contains points where the sums of the distances of this point from the first focus and that from the second focus will always be the same.

This means that as long as $P$ is part of the conic section’s curve, $\overline{PF_1} + \overline{PF_2} = k$, where $k$ is a constant. The center of the ellipse is located right between the two foci. Here are a few more properties of an ellipse that we need to keep in mind:

• The ellipses we’ll work with are either oriented horizontally or vertically.
• The longer segment that passes through one vertex to another is called the major axis.
• The shorter segment would then be the ellipse’s minor axis. The image above shows the two types of ellipses – one with a vertical major axis and the second with a horizontal major axis.

Make sure to familiarize yourself with the terms and labels used here since we’ll need to refer to these components in the later section to help if you know these by heart.

What if we’re only given the graph or some components of the ellipse? Will we be able to find the equation that represents it? The next section will answer these questions.

## How to find the equation of an ellipse?

There are different ways for us to determine the equation of an ellipse – and this will also depend on what’s given for us to work on. One significant component of ellipses is their centers.

Ellipses may be centered at the origin or a particular coordinate, $(h, k)$. This translation will affect the ellipse’s formula.  What we can do is summarize the standard forms of the ellipses into four categories.

Ellipse Formula

Beginning with the standard forms of ellipses that are centered at the origin.

 When the Major Axis is Horizontal When the Major Axis is Vertical \begin{aligned} \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\end{aligned} \begin{aligned} \dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1\end{aligned}

For the equations, the major axis is $2a$ long, so half of the axis will reflect the value of $a$. The value of $b$ represents half the minor axis, so its length will be $2b$. These are the two forms of ellipses centered at the origin. Both show ellipses with major axes that are $2a$ units long and minor axes that are $2b$ units long.

As we can see from the graphs, the foci’s positions will depend on the major axis orientation.

What happens when we translate the ellipse $h$ units to the right and $k$ units vertically? From the quick graph shown above, we can see that the center, foci, and vertices will be translated. What happens with the equation of the newly translated ellipse?

Here are the general forms of ellipses that are centered at $(h, k)$.

 When the Major Axis is Horizontal When the Major Axis is Vertical \begin{aligned} \dfrac{(x – h)^2}{a^2} + \dfrac{(y – k)^2}{b^2} = 1\end{aligned} \begin{aligned} \dfrac{(x -h)^2}{b^2} + \dfrac{(y – k)^2}{a^2} = 1\end{aligned}

The major axis’ length will still be $2a$ long and the ellipses’ minor axis will also remain $2b$ units. As can be seen from the two graphs, these exhibit similar components with the ellipses centered at the origin.

• When the major axis is parallel to the $x$-axis, the foci and vertices will be translated $h$ units to the right and left.
• When the major axis is parallel to the $y$-axis, the foci and vertices will be translated $k$ units upward and downward.

This means that when we know the center of the ellipse, it will be easy for us to find the other components.

### How to find the center of an ellipse?

The center of the ellipse is found right in the middle of the ellipse, hence, its name. There are different ways for us to find the ellipse’s center depending on what’s given.

Finding the Center Given the Equation

As we have shown in the previous section, the center of an ellipse can either be the origin or a particular coordinate, $(h, k)$.

• When given the equation, if the equation is the standard form and only contains $x^2$ and $y^2$ on the numerator, the center will be $(0, 0)$.
• When given an equation in standard and the numerators are $(x-h)^2$ and $(y – k)^2$, the center of the ellipse will be found at $(h, k)$.

Finding the Center Given the Foci or Vertices

What if we’re only given the foci or the vertices of the ellipse instead? Use the fact that the center of the ellipse is simply the midpoint between the foci or the vertices.

Use the midpoint formula, $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$, to find the coordinate of the center.

• This means that when given the foci, find the average of foci’s $x$ and $y$ coordinates.
• Apply a similar process for the major axis or minor axis vertices, depending on what’s given.

### How to find the foci of an ellipse?

We can determine the foci’s distance from the center using the equation’s denominator’s values, $a^2$ and $b^2$.

\begin{aligned}c^2 &= a^2 – b^2 \\ c&= \sqrt{a^2 – b^2}\end{aligned}

Once we have the value of $c$, we can determine the foci (or focus in singular).

• When given the center, we move $c$ units vertically upward or downward when the major axis is parallel to the $y$-axis.
• When given the center, we move $c$ units vertically to the left and right when the major axis is parallel to the $x$-axis.
 Center $(0,0)$ $(h, k)$ Focus when major axis is horizontal $(-c, 0), (c, 0)$ $(h – c, k), (h + c, k)$ Focus when the major axis is vertical $(0, -c), (0, c)$ $(h, k – c), (h, k + c)$

Now, let’s go ahead and review our knowledge about major and minor axes and apply them to determine the vertices of an ellipse.

### How to find the vertices of an ellipse?

Recall that all ellipses will have a major axis that is $2a$ units long. This length is essential since the vertices represent the endpoints of the ellipse’s major axis.

• Determine the larger value between the denominators of an ellipse’s equation in standard form.
• Once you have $a^2$, find the absolute value of $a$.
• Find the vertices’ positions by translating the points $a$ units to the left and to the right if the major axis is horizontal.
• If the major axis is vertical, move $a$ units upward and downward starting from the center.
 Center $(0,0)$ $(h, k)$ Vertex when the major axis is horizontal $(-a, 0), (a, 0)$ $(h – a, k), (h + a, k)$ Vertex when the major axis is vertical $(0, -a), (0, a)$ $(h, k – a), (h, k + a)$

## How to graph an ellipse?

Now that we understand how the different components of an ellipse’s equation affect its graph, here are some important reminders when graphing these ellipses:

• Determine the coordinate of the center, then plot the center.
• Inspect the values of $a^2$ and $b^2$ to determine if the ellipse is oriented horizontally or vertically.
• Find the foci of the ellipse using the equation, $c^2 = b^2 – a^2$. Each focus must be $c$ units from the center and along the major axis.
• Use the value of $a$ and $b$ to note the ellipse’s minor and major axes.
• Connect the essential points to graph the ellipse’s curve,

Why don’t we go ahead and use these steps to graph the ellipse, $\dfrac{(x- 1)^2}{25} + \dfrac{(y + 2)^2}{9} = 1$?

Starting with the center of the ellipse, we have $h = 1$ and $k = -2$, so it’s $(1, -2)$.

Since $25 > 9$, the ellipse is oriented horizontally with $a^2 = 25$ and $b^2 =9$. We can use these values to find the distance of the focus, $c$.

\begin{aligned} a^2 &=25\\ b^2 &= 9\\\\c^2 &= 25 – 9\\&=16\\c&=4 \end{aligned}

This means that we can locate each of the focus since each is $4$ units away from the center. The graph is also expected to have vertices that are $5$ units away from the center. Make sure that these points are along the major axis.

Now that we have the ellipse’s important components let’s connect the outer points to graph the ellipse. We’ll apply a similar process when graphing other ellipses. Make sure to review your notes before working on the problems that follow this section.

Example 1

Apply what you’ve learned about ellipses and their properties to fill in the blanks and make the following statements true.

a. Given an ellipse’s equation in standard form, $\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1$, $a^2$ is equal to _____________ and $b^2$ is equal to ____________.
b. This means that the $x$-intercepts of the ellipse are ____________ and ___________ while the $y$-intercepts are _______________ and _____________.
c. The foci of the ellipse are located at ______________ and ____________.
d. When graphed, the ellipse will be oriented ___________________.

Solution

The larger valued denominator will represent $a^2$ while the smaller valued denominator will represent $b^2$. This means that $a^2$ is equal to $25$ while $b^2$ is equal to $16$.

We can take the square roots of $a^2$ and $b^2$, and these will represent the intercepts of the graph since the ellipse is centered at the origin.

• The major axis is found in the $y$-axis, so $(0, a)$ and $(0, -a)$ will represent the $y$-intercepts.
• Similarly, the $x$-axis will be represented by the intercepts $(b, 0)$ and $(-b, 0)$.
 Intercepts Coordinates $\boldsymbol{x}$-intercepts \begin{aligned}b^2 &= 16\\b&= 4\\\\x_\text{int}&= (-4,0), (4,0)\end{aligned} $\boldsymbol{y}$-intercepts \begin{aligned}a^2 &= 25\\a&= 5\\\\y_\text{int}&= (0, -5), (0,5)\end{aligned}

To find the foci, let’s begin by finding the value of $c^2$ using $c^2 = a^2 – b^2$.

\begin{aligned}c^2 &= a^2 – b^2\\&= 25 – 16\\&= 9\end{aligned}

Since $c^2 = 9$, this means that $c$ is equal to $3$ and consequently, the foci are located at $(0, -3)$ and $(0, 3)$.

Since the larger denominator is found below $y^2$, the ellipse will have a major axis along the $y$-axis, and consequently, the ellipse is expected to be oriented vertically.

Example 2

Determine the standard form of the equation for the ellipses shown below and locate its foci as well. Solution

Let’s begin with the left ellipse: we can see that this ellipse is centered at the origin, so its standard form can only either be $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} =1$ or $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} =1$.

Since the ellipse’s major axis is parallel to the $x$-axis, $a^2$ should be found right below $x^2$. This eliminates the second option for us.

To find the value of $a$ and $b$, let’s inspect the vertices of the ellipse.

• The $x$-intercepts will represent the major axis’ vertices, $(-a, 0)$ and $(a, 0)$.
• Similarly, we find the value of $b$ by inspecting the $y$-intercepts , $(0, -b)$ and $(0, -b)$.

From inspection, we can see that $a = 5$ and $b = 3$. We can now determine the foci of the ellipse. Since the major axis lies along the $x$-axis, we expect the foci to lie along the $x$-axis as well.

\begin{aligned}c^2 &= a^2 – b^2\\&= 25 – 9\\&= 16\\c&= 4\\\\\text{foci} &= (-4, 0), (4, 0) \end{aligned}

We can also now write the equation of the ellipse since we have $a^2 = 25$ and $b^2 = 9$.

\begin{aligned}\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} &=1\\\dfrac{x^2}{25} + \dfrac{y^2}{9} &=1\end{aligned}

a. This means that the ellipse’s equation is $\dfrac{x^2}{25} + \dfrac{y^2}{9} =1$ and it has foci at $(-4, 0)$ and $(4, 0)$.

For the second ellipse, we can see that it’s centered at $(4, 1)$ and the ellipse is oriented vertically so that the equation will be of the form $\dfrac{(x – h)^2}{b^2} + \dfrac{(y – k)^2}{a^2}$.

• The vertices are $(4, 6)$ and $(4, -4)$, so counting the distances from the center, we can see that $a = 5$.
• Similarly, the endpoints of the major axis are $(2, 1)$ and $(6, 1)$, so $b =2$.

To find the foci, we can also use $c^2 = a^2 – b^2$. Substituting $a = 5$ and $b = 2$, we have $c^2 = 21$. This means that the foci will be $\sqrt{21}$ units above and below the center.

\begin{aligned} \text{foci} = (4, 1 -\sqrt{21}), (4, 1 +\sqrt{21})\end{aligned}

To determine the ellipse’s equation, we simply substitute $(h, k) = (4,1)$, $a^2 = 25$, and $b^2 = 4$ into the standard form.

\begin{aligned}\dfrac{(x – h)^2}{b^2} + \dfrac{(y – k)^2}{a^2}\\&= \dfrac{(x – 4)^2}{4} + \dfrac{(y – 1)^2}{25}\end{aligned}

b. Hence, the ellipse’s equation is $\dfrac{(x – 4)^2}{4} + \dfrac{(y – 1)^2}{25}$ and it has foci at $(4, 1 -\sqrt{21})$ and $(4, 1 +\sqrt{21})$.

Example 3

What is the equation (in standard form) of an ellipse that has the following components?

• A minor axis with endpoints at $(-1, -2)$ and $(9, -2)$.
• A major axis with endpoints at $(4, -8)$ and $(4, 4)$.

Solution

Why don’t we go ahead and plot these endpoints and connect them to form the ellipse? This way, it’ll be easier for us to visualize the ellipse and eventually find the equation representing it. Since the endpoints of the minor and major axes are not $x$ and $y$-intercepts, we’re expecting the graph not to be centered at the origin. This is supported by the graph shown above.

Hence, its equation can either be of the forms $\dfrac{(x- h)^2}{a^2} + \dfrac{(y –k)^2}{b^2} = 1$ or $\dfrac{(x- h)^2}{b^2} + \dfrac{(y –k)^2}{a^2} = 1$.

Let’s begin by finding the center’s coordinate by finding the midpoint between the endpoints (you can use either the minor axis’ endpoints or the major axis’). We’re using the minor axis’ endpoints in the calculation shown below.

\begin{aligned}(h, k)&= \left(\dfrac{-1 + 9}{2}, \dfrac{-2 + -2}{2}\right)\\&= (4, -2)\end{aligned}

Now, let’s find the distances between the endpoints to find the minor axis’s lengths and major axis. We can then determine $a^2$ and $b^2$.

 Length of the Axes Denominator of the Ellipse’s Equation Major Axis’ Endpoints \begin{aligned}2b &= |4 – (-8)| \\&= 12\end{aligned} \begin{aligned} a&= \dfrac{12}{2}\\&=6\\a^2 &= 36\end{aligned} Minor Axis’ Endpoints \begin{aligned}2b &= |9 – (-1)| \\&= 10\end{aligned} \begin{aligned} b&= \dfrac{10}{2}\\&=5\\b^2 &= 25\end{aligned}

Since the major axis is parallel to the $y$-axis, we’ll use the form, $\dfrac{(x- h)^2}{b^2} + \dfrac{(y -k)^2}{a^2} = 1$. Let’s substitute the following values into the equation:

• $(h, k) = (4, -2)$
• $a^2 = 36$
• $b^2 = 25$

\begin{aligned} \dfrac{(x- h)^2}{b^2} + \dfrac{(y –k)^2}{a^2} &= 1\\ \dfrac{(x – 4)^2}{25} + \dfrac{(y + 2)^2}{36}\end{aligned}

This means that the ellipse with the given endpoints will have an equation (in standard form) of $\dfrac{(x – 4)^2}{25} + \dfrac{(y + 2)^2}{36}$.

### Practice Questions

1. Apply what you’ve learned about ellipses and their properties to fill in the blanks and make the following statements true.
a. Given an ellipse’s equation in standard form, $\dfrac{x^2}{36} + \dfrac{y^2}{26} = 1$, $a^2$ is equal to _____________ and $b^2$ is equal to ____________.
b. This means that the $x$-intercepts of the ellipse are ____________ and ___________ while the $y$-intercepts are _______________ and _____________.
c. The foci of the ellipse are located at ______________ and ____________.
d. When graphed, the ellipse will be oriented ___________________.
2. Determine the standard form of the equation for the ellipses shown below and locate its foci as well. 3. What is the equation (in standard form) of an ellipse with the following components?
• A minor axis with endpoints at $(-1, -2)$ and $(9, -2)$.
• A major axis with endpoints at $(4, -8)$ and $(4, 4)$.
4. Graph the ellipse being represented by the equation, $\dfrac{(x +2)^2}{4} + \dfrac{(y – 2)^2}{16} = 1$. Include the vertices and foci on the graph.
5. The equation of an ellipse is $2x^2 + 8y^2 + 20x – 16y + 26 = 0$. Rewrite the equation in standard form then find its foci and graph the ellipse.

1.
a. $36$; $26$
b. $(-6, 0)$ and $(6, 0)$; $(0,-\sqrt{26})$ and $(0, \sqrt{26})$
c.$(-\sqrt{10}, 0)$ and $(\sqrt{10}, 0)$
d. along the $x$- axis
2.
a. Equation: $\dfrac{x^2}{16} + \dfrac{y^2}{36} = 1$
Foci: $(-2\sqrt{5}, 0)$ and $(2\sqrt{5}, 0)$
b. Equation: $\dfrac{(x – 2)^2}{64} + \dfrac{(y + 1)^2}{16} = 1$
Foci: $(2- 4\sqrt{3}, -1)$ and $(2+ 4\sqrt{3}, -1)$

3. $\dfrac{(x – 4)^2}{100} + \dfrac{(y + 2)^2}{144} = 1$

4. Foci: $(-2, 2-2\sqrt{3})$ and $(-2, 2 +2\sqrt{3})$ 5. Equation: $\dfrac{(x +5)^2}{16} + \dfrac{(y – 1)^2}{4} = 1$

Foci: $(-5-2\sqrt{3}, 1)$ and $(-5 + 2\sqrt{3}, 1)$ Images/mathematical drawings are created with GeoGebra.