Chapter 9.2, Problem 72E

To determine

To find the equation of the ellipse.

Answer to Problem 72E

The equation of the ellipse is $\frac{{\left(x-6\right)}^2}{324}+\frac{{\left(y-2\right)}^2}{308}=1$ .

Explanation of Solution

Given information:

The given statement is:

“Find the equation of an ellipse such that for any point on the ellipse, the sum of the distances from the points $\left(2,2\right)\text{ and }\left(10,2\right)$ is $36$ .”

Formula used:

The following formula is used:

  $b^2=a^2-c^2$

Calculation:

The two points $\left(2,2\right)\text{ and }\left(10,2\right)$ are the foci and the sum of distance is $2a$ .

By the location of the foci, it is seen that the ellipse has a major horizontal axis.

The center of the ellipse is the mid-point of the foci:

  $\left(h,k\right)=\left(\frac{2+10}{2},\frac{2+2}{2}\right)=\left(6,2\right)$

The length of the major axis is $2a=36$ , calculate the value of $a$ :

  $\begin{array}{l}2a=36\\ a=18\end{array}$

The distance from the center and foci is $c$ :

  $\begin{array}{l}c=\left|10-6\right|=\left|2-6\right|=4\\ c=4\end{array}$

Calculate the value of $b$ :

  $\begin{array}{l}{b}^2=a^2-c^2\\ b^2={18}^2-4^2\\ b^2=308\end{array}$

The equation of the ellipse is:

  $\begin{array}{l}\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1\\ \frac{{\left(x-6\right)}^2}{{18}^2}+\frac{{\left(y-2\right)}^2}{308}=1\\ \frac{{\left(x-6\right)}^2}{324}+\frac{{\left(y-2\right)}^2}{308}=1\end{array}$

Hence, the equation of the ellipse is $\frac{{\left(x-6\right)}^2}{324}+\frac{{\left(y-2\right)}^2}{308}=1$ .