Ellipses

By the end of this section, you will be able to:

Graph an ellipse with center at the origin
Find the equation of an ellipse with center at the origin
Graph an ellipse with center not at the origin
Solve application with ellipses

Graph an Ellipse with Center at the Origin

The next conic section we will look at is an ellipse. We define an ellipse as all points in a plane where the sum of the distances from two fixed points is constant. Each of the given points is called a focus of the ellipse.

Ellipse

An ellipse is all points in a plane where the sum of the distances from two fixed points is constant. Each of the fixed points is called a focus of the ellipse.

We can draw an ellipse by taking some fixed length of flexible string and attaching the ends to two thumbtacks. We use a pen to pull the string taut and rotate it around the two thumbtacks. The figure that results is an ellipse.

A line drawn through the foci intersect the ellipse in two points. Each point is called a vertex of the ellipse. The segment connecting the vertices is called the major axis. The midpoint of the segment is called the center of the ellipse. A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the minor axis.

We mentioned earlier that our goal is to connect the geometry of a conic with algebra. Placing the ellipse on a rectangular coordinate system gives us that opportunity. We place the ellipse so the foci $(-c,0),(c,0)$ are on the $x$ -axis and the center is the origin.

The definition states the sum of the distance from the foci to a point $(x,y)$ is constant. So $d_1+d_2$ is a constant that we will call $2a$ , so $d_1+d_2=2a$ . We will use the distance formula to lead us to an algebraic formula for an ellipse.

Use the distance formula to find $d_1,d_2$ :

\sqrt{(x-(-c))^2+(y-0)^2}+\sqrt{(x-c)^2+(y-0)^2}=2a.

After eliminating radicals and simplifying, we get

\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1.

To simplify the equation of the ellipse, we let $a^2-c^2=b^2$ . So, the equation of an ellipse centered at the origin in standard form is

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.

To graph the ellipse, it will be helpful to know the intercepts. We will find the $x$ -intercepts and $y$ -intercepts using the formula.

$y$ -intercepts	$x$ -intercepts
Let $x=0$ . Then $\tfrac{0^2}{a^2}+\tfrac{y^2}{b^2}=1$ , so $y^2=b^2$ and $y=\pm b$ . The $y$ -intercepts are $(0,b)$ and $(0,-b)$ .	Let $y=0$ . Then $\tfrac{x^2}{a^2}+\tfrac{0^2}{b^2}=1$ , so $x^2=a^2$ and $x=\pm a$ . The $x$ -intercepts are $(a,0)$ and $(-a,0)$ .

Standard Form of the Equation of an Ellipse with Center $(0,0)$

The standard form of the equation of an ellipse with center $(0,0)$ is

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.

The $x$ -intercepts are $(a,0)$ and $(-a,0)$ .

The $y$ -intercepts are $(0,b)$ and $(0,-b)$ .

Notice that when the major axis is horizontal, the value of $a$ will be greater than the value of $b$ and when the major axis is vertical, the value of $b$ will be greater than the value of $a$ . We will use this information to graph an ellipse that is centered at the origin.

Ellipse with center $(0,0)$	$\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$ , $a>b$	$\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$ , $b>a$
Major axis	on the $x$ -axis	on the $y$ -axis
$x$ -intercepts	$(-a,0),(a,0)$	$(-a,0),(a,0)$
$y$ -intercepts	$(0,-b),(0,b)$	$(0,-b),(0,b)$

Example 11.20

How to Graph an Ellipse with Center $(0,0)$ . Graph

\frac{x^2}{4}+\frac{y^2}{9}=1.

Solution.

The equation is in standard form.
Since $9>4$ and $9$ is in the $y^2$ -term, the major axis is vertical.
The endpoints will be the $y$ -intercepts. Since $b^2=9$ , then $b=\pm3$ . The endpoints of the major axis are $(0,3),(0,-3)$ .
The endpoints will be the $x$ -intercepts. Since $a^2=4$ , then $a=\pm2$ . The endpoints of the minor axis are $(2,0),(-2,0)$ .
Sketch the ellipse through the four intercepts.

For $\tfrac{x^2}{4}+\tfrac{y^2}{16}=1$ , enter the positive $y$ -coordinate of a vertex.

For $\tfrac{x^2}{9}+\tfrac{y^2}{16}=1$ , enter the positive $y$ -coordinate of a vertex.

We summarize the steps for reference.

How To: Graph an Ellipse with Center $(0,0)$

Write the equation in standard form.
Determine whether the major axis is horizontal or vertical.
Find the endpoints of the major axis.
Find the endpoints of the minor axis.
Sketch the ellipse.

Sometimes our equation will first need to be put in standard form.

Example 11.21

Graph $x^2+4y^2=16$ .

Solution. We recognize this as the equation of an ellipse since both the $x$ and $y$ terms are squared and have different coefficients.

To get the equation in standard form, divide both sides by 16 so that the equation is equal to 1.

\begin{aligned} x^2+4y^2&=16\\ \frac{x^2}{16}+\frac{4y^2}{16}&=\frac{16}{16}\\ \frac{x^2}{16}+\frac{y^2}{4}&=1. \end{aligned}

The equation is in standard form. The ellipse is centered at the origin, so the center is $(0,0)$ . Since $16>4$ and 16 is in the $x^2$ -term, the major axis is horizontal.

Since $a^2=16$ , $a=\pm4$ , and since $b^2=4$ , $b=\pm2$ . The vertices are $(4,0),(-4,0)$ . The endpoints of the minor axis are $(0,2),(0,-2)$ . Sketch the ellipse.

Write $9x^2+16y^2=144$ in standard form.

Write $16x^2+25y^2=400$ in standard form.

Find the Equation of an Ellipse with Center at the Origin

If we are given the graph of an ellipse, we can find the equation of the ellipse.

Example 11.22

Find the equation of the ellipse centered at $(0,0)$ , with vertices $(-4,0),(4,0)$ and endpoints of the minor axis $(0,3),(0,-3)$ .

Solution. We recognize this as an ellipse that is centered at the origin:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.

Since the major axis is horizontal and the distance from the center to the vertex is 4, we know $a=4$ and so $a^2=16$ :

\frac{x^2}{16}+\frac{y^2}{b^2}=1.

The minor axis is vertical and the distance from the center to the ellipse is 3, so we know $b=3$ and $b^2=9$ :

\frac{x^2}{16}+\frac{y^2}{9}=1.

Find the equation of the ellipse centered at $(0,0)$ with $x$ -intercepts $(-2,0),(2,0)$ and $y$ -intercepts $(0,-5),(0,5)$ .

Find the equation of the ellipse centered at $(0,0)$ with $x$ -intercepts $(-3,0),(3,0)$ and $y$ -intercepts $(0,-2),(0,2)$ .

Graph an Ellipse with Center Not at the Origin

The ellipses we have looked at so far have all been centered at the origin. We will now look at ellipses whose center is $(h,k)$ .

The equation is $\tfrac{(x-h)^2}{a^2}+\tfrac{(y-k)^2}{b^2}=1$ and when $a>b$ , the major axis is horizontal so the distance from the center to the vertex is $a$ . When $b>a$ , the major axis is vertical so the distance from the center to the vertex is $b$ .

Standard Form of the Equation of an Ellipse with Center $(h,k)$

The standard form of the equation of an ellipse with center $(h,k)$ is

\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1.

When $a>b$ , the major axis is horizontal so the distance from the center to the vertex is $a$ .

When $b>a$ , the major axis is vertical so the distance from the center to the vertex is $b$ .

Example 11.23

Graph

\frac{(x-3)^2}{9}+\frac{(y-1)^2}{4}=1.

Solution. The equation is in standard form. The ellipse is centered at $(h,k)$ , so the center is $(3,1)$ . Since $9>4$ and 9 is in the $x^2$ -term, the major axis is horizontal.

Since $a^2=9$ , $a=\pm3$ , and since $b^2=4$ , $b=\pm2$ . The distance from the center to the vertices is 3. The distance from the center to the endpoints of the minor axis is 2. The vertices are $(0,1),(6,1)$ and the endpoints of the minor axis are $(3,3),(3,-1)$ . Sketch the ellipse.

For $\tfrac{(x+3)^2}{4}+\tfrac{(y-5)^2}{16}=1$ , enter the positive distance from the center to a vertex.

For $\tfrac{(x-1)^2}{25}+\tfrac{(y+3)^2}{16}=1$ , enter the positive distance from the center to a vertex.

If we look at the equations $\tfrac{x^2}{9}+\tfrac{y^2}{4}=1$ and $\tfrac{(x-3)^2}{9}+\tfrac{(y-1)^2}{4}=1$ , we see that they are both ellipses with $a=3$ and $b=2$ . So they will have the same size and shape. They are different in that they do not have the same center.

Notice that we could have graphed $\tfrac{(x-3)^2}{9}+\tfrac{(y-1)^2}{4}=1$ by translations. We moved the original ellipse to the right 3 units and then up 1 unit.

In the next example we will use the translation method to graph the ellipse.

Example 11.24

Graph

\frac{(x+4)^2}{16}+\frac{(y-6)^2}{9}=1

by translation.

Solution. This ellipse will have the same size and shape as $\tfrac{x^2}{16}+\tfrac{y^2}{9}=1$ , whose center is $(0,0)$ . We graph this ellipse first.

Since $16>9$ , the major axis is horizontal. Since $a^2=16$ , $a=\pm4$ , and since $b^2=9$ , $b=\pm3$ . The vertices are $(4,0),(-4,0)$ . The endpoints of the minor axis are $(0,3),(0,-3)$ . Sketch the ellipse.

The original equation is in standard form,

\frac{(x-(-4))^2}{16}+\frac{(y-6)^2}{9}=1.

The ellipse is centered at $(h,k)$ , so the center is $(-4,6)$ . We translate the graph of $\tfrac{x^2}{16}+\tfrac{y^2}{9}=1$ four units to the left and then up 6 units. Verify that the center is $(-4,6)$ . The new ellipse is the ellipse whose equation is

\frac{(x+4)^2}{16}+\frac{(y-6)^2}{9}=1.

For $\tfrac{(x-5)^2}{9}+\tfrac{(y+4)^2}{4}=1$ , enter the $x$ -coordinate of the center.

For $\tfrac{(x+6)^2}{16}+\tfrac{(y+2)^2}{25}=1$ , enter the $y$ -coordinate of the center.

When an equation has both an $x^2$ and a $y^2$ with different coefficients, we verify that it is an ellipse by putting it in standard form. We will then be able to graph the equation.

Example 11.25

Write the equation

x^2+4y^2-4x+24y+24=0

in standard form and graph.

Solution. We put the equation in standard form by completing the squares in both $x$ and $y$ .

\begin{aligned} x^2+4y^2-4x+24y+24&=0\\ (x^2-4x+\underline{\phantom{4}})+ (4y^2+24y+\underline{\phantom{36}})&=-24\\ (x^2-4x+\underline{\phantom{4}})+ 4(y^2+6y+\underline{\phantom{9}})&=-24\\ (x^2-4x+4)+4(y^2+6y+9)&=-24+4+36\\ (x-2)^2+4(y+3)^2&=16\\ \frac{(x-2)^2}{16}+\frac{4(y+3)^2}{16}&=\frac{16}{16}\\ \frac{(x-2)^2}{16}+\frac{(y+3)^2}{4}&=1. \end{aligned}

The equation is in standard form. The ellipse is centered at $(h,k)$ , so the center is $(2,-3)$ . Since $16>4$ and 16 is in the $x^2$ -term, the major axis is horizontal. Since $a^2=16$ , $a=\pm4$ , and since $b^2=4$ , $b=\pm2$ . The distance from the center to the vertices is 4. The distance from the center to the endpoints of the minor axis is 2. The vertices are $(-2,-3),(6,-3)$ and the endpoints of the minor axis are $(2,-1),(2,-5)$ . Sketch the ellipse.

Write $6x^2+4y^2+12x-32y+34=0$ in standard form.

Write $4x^2+y^2-16x-6y+9=0$ in standard form.

Solve Application with Ellipses

The orbits of the planets around the sun follow elliptical paths.

Example 11.26

Pluto (a dwarf planet) moves in an elliptical orbit around the Sun. The closest Pluto gets to the Sun is approximately 30 astronomical units (AU) and the furthest is approximately 50 AU. The Sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit has vertices $(-40,0),(40,0)$ and the Sun at $(10,0)$ . Use the graph to write an equation for the elliptical orbit of Pluto.

Solution. We recognize this as an ellipse that is centered at the origin:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.

Since the major axis is horizontal and the distance from the center to the vertex is 40, we know $a=40$ and so $a^2=1600$ :

\frac{x^2}{1600}+\frac{y^2}{b^2}=1.

The minor axis is vertical but the end points aren’t given. To find $b$ we will use the location of the Sun. Since the Sun is a focus of the ellipse at the point $(10,0)$ , we know $c=10$ . Use this to solve for $b^2$ :

\begin{aligned} b^2&=a^2-c^2\\ b^2&=40^2-10^2\\ b^2&=1600-100\\ b^2&=1500. \end{aligned}

Substitute $a^2$ and $b^2$ into the standard form of the ellipse:

\frac{x^2}{1600}+\frac{y^2}{1500}=1.

A planet's elliptical orbit has vertices $(-25,0),(25,0)$ and its sun is a focus at $(5,0)$ . Write the equation of the orbit.

A planet's elliptical orbit has vertices $(-35,0),(35,0)$ and its sun is a focus at $(15,0)$ . Write the equation of the orbit.

Key terms. An ellipse is all points in a plane where the sum of the distances from two fixed points is constant. Each fixed point is a focus. The line through the foci is the major axis, its endpoints are the vertices, the midpoint of the vertices is the center, and the perpendicular line through the center is the minor axis.

This section is adapted from Intermediate Algebra 2e, Section 11.3 by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at OpenStax. Changes: omitted readiness quizzes, practice sets, self-checks, media links, and complex source figures; converted Try It problems to interactive questions and described source figures in words.

Parabolas Hyperbolas