Chapter 9.2, Problem 26E

To determine

To find:The standard equation of the ellipse when foci are $\left(0,0\right)$ and $\left(4,0\right)$ and the length of major axis is $6$ .

Answer to Problem 26E

The equation of the ellipse is $\frac{{\left(x-2\right)}^2}{9}+\frac{y^2}{5}=1$ .

Explanation of Solution

Given information:

The foci of the ellipse are $\left(0,0\right)$ , $\left(4,0\right)$ and length of major axis is $6$ .

Calculation:

The foci of the ellipse are $\left(0,0\right)$ , $\left(4,0\right)$ . Since the foci of the ellipse lies on thex-axis. The major axis is horizontal. So, the general equation of the ellipse is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ .

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

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

So, the center of the ellipse is the point $\left(2,0\right)$ .

The length of the major axis is $6$ .

Use the formula of length of major axis.

  $\begin{array}{c}2a=6\\ a=3\end{array}$

Use the formula for $c$ .

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

Substitute $3$ for $a$ and $2$ for $c$ in the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ 2^2=3^2-b^2\\ b^2=9-4\\ b^2=5\\ b=\sqrt{5}\end{array}$

Substitute $3$ for $a$ , $\sqrt{5}$ for $b$ , $2$ for $h$ and $0$ for $k$ in the general equation of ellipse.

  $\begin{array}{c}\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1\\ \frac{{\left(x-2\right)}^2}{3^2}+\frac{{\left(y-0\right)}^2}{{\left(\sqrt{5}\right)}^2}=1\\ \frac{{\left(x-2\right)}^2}{9}+\frac{y^2}{5}=1\end{array}$

Therefore, the equation of the ellipse is $\frac{{\left(x-2\right)}^2}{9}+\frac{y^2}{5}=1$ .