Chapter 9.2, Problem 28E

To determine

To find:The standard equation of the ellipse when center is $\left(0,4\right)$ , $a=5c$ and vertices are $\left(0,-1\right)$ and $\left(0,9\right)$ .

Answer to Problem 28E

The equation of the ellipse is $\frac{x^2}{6}+\frac{4y^2}{25}=1$ .

Explanation of Solution

Given information:

The vertices of the ellipse are $\left(0,-1\right)$ , $\left(0,9\right)$ , $a=5c$ andthe centeris $\left(0,4\right)$ .

Calculation:

The vertices of the ellipse are $\left(0,-1\right)$ and $\left(0,9\right)$ . 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 the point $\left(0,4\right)$ .

The length of the major axis is half the distance between points $\left(0,-1\right)$ and $\left(0,9\right)$ .

  $\begin{array}{c}\frac{1}{2}\left[\sqrt{{\left(0-0\right)}^2+{\left(9-\left(-1\right)\right)}^2}\right]=\frac{1}{2}\left[\sqrt{{10}^2}\right]\\ =\frac{1}{2}\left(10\right)\\ =5\end{array}$

Use the formula of length of major axis.

  $\begin{array}{c}2a=5\\ a=\frac{5}{2}\end{array}$

Use the formula for $c$ .

  $\begin{array}{c}a=5c\\ c=\frac{a}{5}\\ c=\frac{1}{2}\end{array}$

Substitute $\frac{5}{2}$ for $a$ and $\frac{1}{2}$ for $c$ in the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ {\left(\frac{1}{2}\right)}^2={\left(\frac{5}{2}\right)}^2-b^2\\ b^2=\frac{25}{4}-\frac{1}{4}\\ b^2=\frac{24}{4}\\ b=\sqrt{6}\end{array}$

Substitute $\frac{5}{2}$ for $a$ , $\sqrt{6}$ for $b$ , $0$ for $h$ and $4$ for $k$ in the general equation of ellipse.

  $\begin{array}{c}\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1\\ \frac{{\left(x-0\right)}^2}{{\left(\sqrt{6}\right)}^2}+\frac{{\left(y-4\right)}^2}{{\left(\frac{5}{2}\right)}^2}=1\\ \frac{x^2}{6}+\frac{4y^2}{25}=1\end{array}$

Therefore, the equation of the ellipse is $\frac{x^2}{6}+\frac{4y^2}{25}=1$ .