Chapter 9.2, Problem 25E

To determine

To find:The standard equation of the ellipse when foci are $\left(0,0\right)$ and $\left(0,8\right)$ and the length of major axis is $36$ .

Answer to Problem 25E

The equation of the ellipse is $\frac{x^2}{308}+\frac{{\left(y-4\right)}^2}{324}=1$ .

Explanation of Solution

Given information:

The foci of the ellipse are $\left(0,0\right)$ , $\left(0,8\right)$ and length of major axis is $36$ .

Calculation:

The foci of the ellipse are $\left(0,0\right)$ , $\left(0,8\right)$ .

As the foci of the ellipse lies on y-axis. So, the major axis is vertical.So, the general equation of the ellipse is $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$ .

The center of the ellipse is at midpoint of the foci.

  $\frac{1}{2}\left[\left(0,0\right)+\left(0,8\right)\right]=\left(0,4\right)$

So, the center of the ellipse is the point $\left(0,4\right)$ .

The length of the major axis is $36$ .

Use the formula of length of major axis.

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

Use the formula for $c$ .

  $\begin{array}{c}c=\frac{1}{2}\left[\sqrt{{\left(0-0\right)}^2+{\left(8-0\right)}^2}\right]\\ =\frac{1}{2}\left[\sqrt{8^2}\right]\\ =\frac{1}{2}\left(8\right)\\ =4\end{array}$

Substitute $18$ for $a$ and $4$ for $c$ in the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ 4^2={18}^2-b^2\\ b^2=324-16\\ b^2=308\\ b=2\sqrt{77}\end{array}$

Substitute $18$ for $a$ , $2\sqrt{77}$ for $b$ , $0$ for $h$ and $4$ for $k$ in the general equation of the ellipse.

  $\begin{array}{c}\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1\\ \frac{{\left(x-0\right)}^2}{{\left(2\sqrt{77}\right)}^2}+\frac{{\left(y-4\right)}^2}{{18}^2}=1\\ \frac{x^2}{308}+\frac{{\left(y-4\right)}^2}{324}=1\end{array}$

Therefore, the equation of the ellipse is $\frac{x^2}{308}+\frac{{\left(y-4\right)}^2}{324}=1$ .