10.1 The ellipse  (Page 7/16)

We must begin by rewriting the equation in standard form.

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

Factor out the coefficients of the squared terms.

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

Rewrite as perfect squares.

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

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 $\left(h,k\right)=\left(5,\mathrm{-2}\right)$
• the coordinates of the vertices are $\left(h\pm a,k\right)=\left(5\pm \sqrt{9},\mathrm{-2}\right)=\left(5\pm 3,\mathrm{-2}\right),$ or $\left(2,\mathrm{-2}\right)$ and $\left(8,\mathrm{-2}\right)$
• the coordinates of the co-vertices are $\left(h,k\pm b\right)=\left(\text{5},\mathrm{-2}\pm \sqrt{4}\right)=\left(\text{5},\mathrm{-2}\pm 2\right),$ or $\left(5,\mathrm{-4}\right)$ 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:

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

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] .

1. 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 $\frac{x^2}{2304}+\frac{y^2}{529}=1.$

2. 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.