Example: Writing the Equation of an Ellipse Centered at a Point Other Than the Origin

What is the standard form equation of the ellipse that has vertices $\left(-2,-8\right)$ and $\left(-2,\text{2}\right)$ and foci $\left(-2,-7\right)$ and $\left(-2,\text{1}\right)?$

Answer: The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form

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

First, we identify the center, $\left(h,k\right)$. The center is halfway between the vertices, $\left(-2,-8\right)$ and $\left(-2,\text{2}\right)$. Applying the midpoint formula, we have:

$\begin{array}{l}\left(h,k\right)=\left(\frac{-2+\left(-2\right)}{2},\frac{-8+2}{2}\right)\hfill \\ \text{ }=\left(-2,-3\right)\hfill \end{array}$

Next, we find ${a}^{2}$. The length of the major axis, $2a$, is bounded by the vertices. We solve for $a$ by finding the distance between the y-coordinates of the vertices.

$\begin{array}{c}2a=2-\left(-8\right)\\ 2a=10\\ a=5\end{array}$

So ${a}^{2}=25$. Now we find ${c}^{2}$. The foci are given by $\left(h,k\pm c\right)$. So, $\left(h,k-c\right)=\left(-2,-7\right)$ and $\left(h,k+c\right)=\left(-2,\text{1}\right)$. We substitute $k=-3$ using either of these points to solve for $c$.

$\begin{array}{c}k+c=1\\ -3+c=1\\ c=4\end{array}$ So ${c}^{2}=16$.

Next, we solve for ${b}^{2}$ using the equation ${c}^{2}={a}^{2}-{b}^{2}$.

$\begin{array}{c}{c}^{2}={a}^{2}-{b}^{2}\\ 16=25-{b}^{2}\\ {b}^{2}=9\end{array}$

Finally, we substitute the values found for $h,k,{a}^{2}$, and ${b}^{2}$ into the standard form equation for an ellipse:

$\frac{{\left(x+2\right)}^{2}}{9}+\frac{{\left(y+3\right)}^{2}}{25}=1$

Example: 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. a. 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 $\left(0,0\right)$. b. 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.

Answer: a. We are assuming a horizontal ellipse with center $\left(0,0\right)$, 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 $\frac{{x}^{2}}{2304}+\frac{{y}^{2}}{529}=1$. b. To find the distance between the senators, we must find the distance between the foci, $\left(\pm c,0\right)$, where ${c}^{2}={a}^{2}-{b}^{2}$. Solving for $c$, we have:

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

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

Example: 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.

Answer: 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 $\left(0,0\right)$
• the coordinates of the vertices are $\left(0,\pm a\right)=\left(0,\pm \sqrt{25}\right)=\left(0,\pm 5\right)$
• the coordinates of the co-vertices are $\left(\pm b,0\right)=\left(\pm \sqrt{9},0\right)=\left(\pm 3,0\right)$
• the coordinates of the foci are $\left(0,\pm c\right)$, where ${c}^{2}={a}^{2}-{b}^{2}$ Solving for $c$, we have:

$\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}}\hfill \\ =\pm \sqrt{25 - 9}\hfill \\ =\pm \sqrt{16}\hfill \\ =\pm 4\hfill \end{array}$

Therefore, the coordinates of the foci are $\left(0,\pm 4\right)$. Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

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

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

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

$\begin{array}{l} 4{x}^{2}+25{y}^{2}=100\hfill \\ \text{ }\frac{4{x}^{2}}{100}+\frac{25{y}^{2}}{100}=\frac{100}{100}\hfill \\ \text{ }\frac{{x}^{2}}{25}+\frac{{y}^{2}}{4}=1\hfill \end{array}$

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 $\left(0,0\right)$
• the coordinates of the vertices are $\left(\pm a,0\right)=\left(\pm \sqrt{25},0\right)=\left(\pm 5,0\right)$
• the coordinates of the co-vertices are $\left(0,\pm b\right)=\left(0,\pm \sqrt{4}\right)=\left(0,\pm 2\right)$
• the coordinates of the foci are $\left(\pm c,0\right)$, where ${c}^{2}={a}^{2}-{b}^{2}$. Solving for $c$, we have:

$\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}}\hfill \\ =\pm \sqrt{25 - 4}\hfill \\ =\pm \sqrt{21}\hfill \end{array}$

Therefore the coordinates of the foci are $\left(\pm \sqrt{21},0\right)$. Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Example: Graphing an Ellipse Centered at (h, k)

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

Answer: 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 $\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1$, where ${b}^{2}=4$ and ${a}^{2}=9$. It follows that:

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

$\begin{array}{l}\begin{array}{l}\\ c=\pm \sqrt{{a}^{2}-{b}^{2}}\end{array}\hfill \\ =\pm \sqrt{9 - 4}\hfill \\ =\pm \sqrt{5}\hfill \end{array}$

Therefore, the coordinates of the foci are $\left(-2,\text{5}-\sqrt{5}\right)$ and $\left(-2,\text{5+}\sqrt{5}\right)$. Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

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

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

Answer: We must begin by rewriting the equation in standard form. $$4{x}^{2}+9{y}^{2}-40x+36y+100=0$$ Group terms that contain the same variable, and move the constant to the opposite side of the equation. $$\left(4{x}^{2}-40x\right)+\left(9{y}^{2}+36y\right)=-100$$ Factor out the coefficients of the squared terms. $$4\left({x}^{2}-10x\right)+9\left({y}^{2}+4y\right)=-100$$ Complete the square twice. Remember to balance the equation by adding the same constants to each side. $$4\left({x}^{2}-10x+25\right)+9\left({y}^{2}+4y+4\right)=-100+100+36$$ Rewrite as perfect squares. $$4{\left(x - 5\right)}^{2}+9{\left(y+2\right)}^{2}=36$$ Divide both sides by the constant term to place the equation in standard form. $$\frac{{\left(x - 5\right)}^{2}}{9}+\frac{{\left(y+2\right)}^{2}}{4}=1$$ 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 $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$, where ${a}^{2}=9$ and ${b}^{2}=4$. It follows that:

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

$$\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}}\hfill \\ =\pm \sqrt{9 - 4}\hfill \\ =\pm \sqrt{5}\hfill \end{array}$$ Therefore, the coordinates of the foci are $\left(\text{5}-\sqrt{5},-2\right)$ and $\left(\text{5+}\sqrt{5},-2\right)$. Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.