Chapter 11, Problem 44RE

To determine

To find: The equation of the ellipse.

Answer to Problem 44RE

The equation of ellipse is

  $\frac{x^2}{9}+\frac{{\left(y-4\right)}^2}{25}=1$

Explanation of Solution

Given information:

An ellipse with:

  $\begin{array}{l}center\left(0,4\right)\\ Foci\left(0,0\right)\&\left(0,8\right)\\ \text{major axis length}=10\end{array}$

Calculation:

  $\begin{array}{l}center\left(h,k\right)\\ c={\text{distance between F}}_1\&C{\text{ or F}}_2\&C\\ c^2=a^2+b^2\end{array}$

Given

  $\begin{array}{l}center=\left(0,4\right)\Rightarrow h=0,k=4\\ Foci=\left(0,0\right)\&\left(0,8\right)\\ c=\sqrt{{\left(0-0\right)}^2+{\left(0-4\right)}^2}\\ c=4\end{array}$

  $\begin{array}{l}\text{Also given that major axis length}=10\\ 2a=10\\ a=5\end{array}$

  $\begin{array}{l}{c}^2=a^2-b^2\\ 4^2=5^2-b^2\\ b^2=25-16\\ b^2=9\Rightarrow b=3\end{array}$

Hence, The equation of ellipse is

  $\begin{array}{l}\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1\\ \frac{{(x-0)}^2}{9}+\frac{{\left(y-4\right)}^2}{25}=1\end{array}$