# 12.1 The ellipse  (Page 7/16)

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## Graphing an ellipse centered at ( h , k ) by first writing it in standard form

Graph the ellipse given by the equation $\text{\hspace{0.17em}}4{x}^{2}+9{y}^{2}-40x+36y+100=0.\text{\hspace{0.17em}}$ Identify and label the center, vertices, co-vertices, and foci.

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 $\text{\hspace{0.17em}}9>4,\text{\hspace{0.17em}}$ the major axis is parallel to the x -axis. Therefore, the equation is in the form $\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}{a}^{2}=9\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{b}^{2}=4.\text{\hspace{0.17em}}$ It follows that:

• the center of the ellipse is $\text{\hspace{0.17em}}\left(h,k\right)=\left(5,-2\right)$
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(h±a,k\right)=\left(5±\sqrt{9},-2\right)=\left(5±3,-2\right),\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\left(2,-2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(8,-2\right)$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(h,k±b\right)=\left(\text{5},-2±\sqrt{4}\right)=\left(\text{5},-2±2\right),\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\left(5,-4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(5,\text{0}\right)$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(h±c,k\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}{c}^{2}={a}^{2}-{b}^{2}.\text{\hspace{0.17em}}$ Solving for $\text{\hspace{0.17em}}c,\text{\hspace{0.17em}}$ we have:
$\begin{array}{l}c=±\sqrt{{a}^{2}-{b}^{2}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=±\sqrt{9-4}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=±\sqrt{5}\hfill \end{array}$

Therefore, the coordinates of the foci are $\text{\hspace{0.17em}}\left(\text{5}-\sqrt{5},-2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(\text{5+}\sqrt{5},-2\right).$

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

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

$4{x}^{2}+{y}^{2}-24x+2y+21=0$

$\text{\hspace{0.17em}}\frac{{\left(x-3\right)}^{2}}{4}+\frac{{\left(y+1\right)}^{2}}{16}=1;\text{\hspace{0.17em}}$ center: $\text{\hspace{0.17em}}\left(3,-1\right);\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(3,-\text{5}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(3,\text{3}\right);\text{\hspace{0.17em}}$ co-vertices: $\text{\hspace{0.17em}}\left(1,-1\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(5,-1\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(3,-\text{1}-2\sqrt{3}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(3,-\text{1+}2\sqrt{3}\right)$

## Solving applied problems involving ellipses

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 [link] . 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.

## 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 [link] .

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 $\text{\hspace{0.17em}}\left(0,0\right).$
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.
1. We are assuming a horizontal ellipse with center $\text{\hspace{0.17em}}\left(0,0\right),$ so we need to find an equation of the form $\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a>b.\text{\hspace{0.17em}}$ We know that the length of the major axis, $\text{\hspace{0.17em}}2a,\text{\hspace{0.17em}}$ is longer than the length of the minor axis, $\text{\hspace{0.17em}}2b.\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}a,$ we have $\text{\hspace{0.17em}}2a=96,$ so $\text{\hspace{0.17em}}a=48,$ and $\text{\hspace{0.17em}}{a}^{2}=2304.$
• Solving for $\text{\hspace{0.17em}}b,$ we have $\text{\hspace{0.17em}}2b=46,$ so $\text{\hspace{0.17em}}b=23,$ and $\text{\hspace{0.17em}}{b}^{2}=529.$

Therefore, the equation of the ellipse is $\text{\hspace{0.17em}}\frac{{x}^{2}}{2304}+\frac{{y}^{2}}{529}=1.$

2. To find the distance between the senators, we must find the distance between the foci, $\text{\hspace{0.17em}}\left(±c,0\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}{c}^{2}={a}^{2}-{b}^{2}.\text{\hspace{0.17em}}$ Solving for $\text{\hspace{0.17em}}c,$ we have:

The points $\text{\hspace{0.17em}}\left(±42,0\right)\text{\hspace{0.17em}}$ represent the foci. Thus, the distance between the senators is $\text{\hspace{0.17em}}2\left(42\right)=84\text{\hspace{0.17em}}$ feet.

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