Graphing Ellipses

Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a > b$ for horizontal ellipses and $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, $a > b$ for vertical ellipses.

HOW TO

Given the standard form of an equation for an ellipse centered at  $(0,0)$,  sketch the graph.

1. Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.
   1. If the equation is in the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,  where $a > b$, then
      • the major axis is the $x$-axis
      • the coordinates of the vertices are $(±a,0)$
      • the coordinates of the co-vertices are $(0,±b)$
      • the coordinates of the foci are $(±c,0)$
   2. If the equation is in the form $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, where $a > b$, then
      • the major axis is the $y$-axis
      • the coordinates of the vertices are $(0,±a)$
      • the coordinates of the co-vertices are $(±b,0)$
      • the coordinates of the foci are $(0,±c)$
2. Solve for $c$ using the equation $c^2=a^2−b^2$.
3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.

EXAMPLE 3

Graphing an Ellipse Centered at the Origin

Graph the ellipse given by the equation, $\frac{x^2}{9}+\frac{y^2}{25}=1$. Identify and label the center, vertices, co-vertices, and foci.

Solution

First, we determine the position of the major axis. Because $25 > 9$, the major axis is on the $y$-axis. Therefore, the equation is in the form $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, where $b^2=9$ and $a^2=25$. It follows that:

• the center of the ellipse is $(0,0)$
• the coordinates of the vertices are $(0,±a)=(0,±\sqrt{25})=(0,±5)$
• the coordinates of the co-vertices are $(±b,0)=(±\sqrt{9},0)=(±3,0)$
• the coordinates of the foci are $(0,±c)$, where $c^2=a^2−b^2$ Solving for $c$, we have:

$\begin{aligned} c &= ± \sqrt{a^2 - b^2} \\ &= ± \sqrt{25-9} \\ &= ± \sqrt{16} \\ &= ±4 \end{aligned}$

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure 8.

Figure 8

TRY IT #3

Graph the ellipse given by the equation $\frac{x^2}{36}+\frac{y^2}{4}=1$. Identify and label the center, vertices, co-vertices, and foci.

EXAMPLE 4

Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form

Graph the ellipse given by the equation $4x^2+25y^2=100$. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

Solution

First, use algebra to rewrite the equation in standard form.

$\begin{aligned} 4x^2 &+25y^2 = 100 \\ \frac{4x^2}{100} &+25y^2 = \frac{100}{100} \\ \frac{x^2}{25} &+ \frac{y^2}{4} = 1 \end{aligned}$

Next, we determine the position of the major axis. Because $25 > 4$, the major axis is on the $x$-axis. Therefore, the equation is in the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a^2=25$ and $b^2=4$. It follows that:

• the center of the ellipse is $(0,0)$
• the coordinates of the vertices are $(±a,0)=(±\sqrt{25},0)=(±5,0)$
• the coordinates of the co-vertices are $(0,±b)=(0, ±\sqrt{4})=(0,±2)$
• the coordinates of the foci are $(±c,0)$, where $c^2=a^2−b^2$ Solving for $c$, we have:

$\begin{aligned} c &= ± \sqrt{a^2 - b^2} \\ &= ± \sqrt{25 - 4} \\ &= ± \sqrt{21} \end{aligned}$

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Figure 9

TRY IT #4

Graph the ellipse given by the equation $49x^2+16y^2=784$.  Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

Graphing Ellipses Not Centered at the Origin

When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, $(h, k)$, we use the standard forms $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1, \quad a > b$ for horizontal ellipses and $\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1, \quad a > b$ for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.

Given the standard form of an equation for an ellipse centered at $(h, k)$, sketch the graph.

1. Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.
   1. If the equation is in the form $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$, where $a > b$, then
      • the center is $(h, k)$
      • the major axis is parallel to the $x$-axis
      • the coordinates of the vertices are $(h \pm a, k)$
      • the coordinates of the co-vertices are $(h, k \pm b)$
      • the coordinates of the foci are $(h \pm c, k)$
   2. If the equation is in the form $\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1$, where $a > b$, then
      • the center is $(h, k)$
      • the major axis is parallel to the $y$-axis
      • the coordinates of the vertices are $(h, k \pm a)$
      • the coordinates of the co-vertices are $(h \pm b, k)$
      • the coordinates of the foci are $(h, k \pm c)$
2. Solve for $c$ using the equation $c^{2}=a^{2}-b^{2}$.
3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.

EXAMPLE 5

Graphing an Ellipse Centered at $(h, k)$

Graph the ellipse given by the equation, $\frac{(x+2)^2}{4} + \frac{(y−5)^2}{9}=1$. Identify and label the center, vertices, co-vertices, and foci.

Solution

First, we determine the position of the major axis. Because $9 > 4$, the major axis is parallel to the $y$ axis. Therefore, the equation is in the form $\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1$, where $b^{2}=4$ and $a^{2}=9$. It follows that:

• the center of the ellipse is $(h, k)=(-2,5)$
• the coordinates of the vertices are $(h, k \pm a)=(-2,5 \pm \sqrt{9})=(-2,5 \pm 3)$, or $(-2,2)$ and $(-2,8)$
• the coordinates of the co-vertices are $(h \pm b, k)=(-2 \pm \sqrt{4}, 5)=(-2 \pm 2,5)$, or $(-4,5)$ and $(0,5)$
• the coordinates of the foci are $(h, k \pm c)$, where $c^{2}=a^{2}-b^{2}$. Solving for $c$, we have:

$\begin{aligned} &c=\pm \sqrt{a^{2}-b^{2}} \\ &=\pm \sqrt{9-4} \\ &=\pm \sqrt{5} \end{aligned}$

Therefore, the coordinates of the foci are $(-2,5-\sqrt{5})$ and $(-2,5+\sqrt{5})$.

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Figure 10

TRY IT #5

Graph the ellipse given by the equation $\frac{(x−4)^2}{36} + \frac{(y−2)^2}{20}=1$. Identify and label the center, vertices, co-vertices, and foci.

HOW TO

Given the general form of an equation for an ellipse centered at $(h, k)$, express the equation in standard form.

1. Recognize that an ellipse described by an equation in the form $a x^{2}+b y^{2}+c x+d y+e=0$ is in general form.
2. Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation.
3. Factor out the coefficients of the $x^{2}$ and $y^{2}$ terms in preparation for completing the square.
4. Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, $m_{1}(x-h)^{2}+m_{2}(y-k)^{2}=m_{3}, w_{h}$ where $m_{1}, m_{2}$, and $m_{3}$ are constants.
5. Divide both sides of the equation by the constant term to express the equation in standard form.

EXAMPLE 6

Graphing an Ellipse Centered at $(h, k)$ by First Writing It in Standard Form

Graph the ellipse given by the equation $4x^2+9y^2−40x+36y+100=0$. Identify and label the center, vertices, co-vertices, and foci.

Solution

$4x^2+9y^2−40x+36y+100=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation.

$(4x^2−40x)+(9y^2+36y)=−100$

Factor out the coefficients of the squared terms.

$4(x^2−10x)+9(y^2+4y)=−100$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$4(x^2−10x+25)+9(y^2+4y+4)=−100+100+36$

Rewrite as perfect squares.

$4(x−5)^2+9(y+2)^2=36$

Divide both sides by the constant term to place the equation in standard form.

$\frac{(x−5)^2}{9} + \frac{(y+2)^2}{4}=1$

Now that the equation is in standard form, we can determine the position of the major axis. Because $9 > 4$, the major axis is parallel to the $x$-axis. Therefore, the equation is in the form $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$, where $a^{2}=9$ and $b^{2}=4$. It follows that:

• the center of the ellipse is $(h, k)=(5,-2)$
• the coordinates of the vertices are $(h \pm a, k)=(5 \pm \sqrt{9},-2)=(5 \pm 3,-2)$, or $(2,-2)$ and $(8,-2)$
• the coordinates of the co-vertices are $(h, k \pm b)=(5,-2 \pm \sqrt{4})=(5,-2 \pm 2)$, or $(5,-4)$ and $(5,0)$
• the coordinates of the foci are $(h \pm c, k)$, where $c^{2}=a^{2}-b^{2}$. Solving for $c$, we have:

$\begin{aligned} &c=\pm \sqrt{a^{2}-b^{2}} \\ &=\pm \sqrt{9-4} \\ &=\pm \sqrt{5} \end{aligned}$

Therefore, the coordinates of the foci are $(5-\sqrt{5},-2)$ and $(5+\sqrt{5},-2)$.

Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse as shown in Figure 11 .

Figure 11

TRY IT #6

Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse.

$4x^2+y^2−24x+2y+21=0$

Source: Rice University, https://openstax.org/books/college-algebra/pages/8-1-the-ellipse
This work is licensed under a Creative Commons Attribution 4.0 License.

Many real-world situations can be represented by ellipses, including orbits of planets, satellites, moons and comets, and shapes of boat keels, rudders, and some airplane wings. A medical device called a lithotripter uses elliptical reflectors to break up kidney stones by generating sound waves. Some buildings, called whispering chambers, are designed with elliptical domes so that a person whispering at one focus can easily be heard by someone standing at the other focus. This occurs because of the acoustic properties of an ellipse. When a sound wave originates at one focus of a whispering chamber, the sound wave will be reflected off the elliptical dome and back to the other focus. See Figure 12. In the whisper chamber at the Museum of Science and Industry in Chicago, two people standing at the foci – about 43 feet apart – can hear each other whisper.

Figure 12 Sound waves are reflected between foci in an elliptical room, called a whispering chamber.

EXAMPLE 7

Locating the Foci of a Whispering Chamber

The Statuary Hall in the Capitol Building in Washington, D.C. is a whispering chamber. Its dimensions are 46 feet wide by 96 feet long as shown in Figure 13.

1. What is the standard form of the equation of the ellipse representing the outline of the room? Hint: assume a horizontal ellipse, and let the center of the room be the point $(0,0)$.
2. If two senators standing at the foci of this room can hear each other whisper, how far apart are the senators? Round to the nearest foot.

Figure 13

Solution

1. We are assuming a horizontal ellipse with center $(0,0)$, so we need to find an equation of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a > b$. We know that the length of the major axis, $2a$, is longer than the length of the minor axis, $2b$. So the length of the room, 96, is represented by the major axis, and the width of the room, 46, is represented by the minor axis.

• Solving for $a$, we have $2a=96$, so $a=48$, and $a^2=2304$.
• Solving for $b$, we have $2b=46$, so $b=23$, and $b^2=529$.

Therefore, the equation of the ellipse is $\frac{x^2}{2304}+\frac{y^2}{529}=1$.

2. To find the distance between the senators, we must find the distance between the foci, $(±c,0)$, where $c^2=a^2−b^2$. Solving for $c$, we have:

$\begin{array}{ll} c^{2}=a^{2}-b^{2} & \\ c^{2}=2304-529 & \text { Substitute using the values found in part (a). } \\ c=\pm \sqrt{2304-529} & \text { Take the square root of both sides. } \\ c=\pm \sqrt{1775} & \text { Subtract. } \\ c \approx \pm 42 & \text { Round to the nearest foot. } \end{array}$

The points $(±42,0)$ represent the foci. Thus, the distance between the senators is $2(42)=84$ feet.

TRY IT #7

Suppose a whispering chamber is 480 feet long and 320 feet wide.

ⓐ What is the standard form of the equation of the ellipse representing the room? Hint: assume a horizontal ellipse, and let the center of the room be the point $(0,0)$.

ⓑ If two people are standing at the foci of this room and can hear each other whisper, how far apart are the people? Round to the nearest foot.