Chapter 9.2, Problem 16E

To determine

To Find: The standard form of equation of ellipse with the given characteristics.

Answer to Problem 16E

The equation of the ellipse is given by $\frac{x^2}{48}+\frac{y^2}{64}=1$

Explanation of Solution

Given information:

  $Vertices=\left(0,\pm 8\right),foci=\left(0,\pm 4\right)$

Concept and Formula Used:

The standard form of an ellipse with center $\left(h,k\right)$ and major axis parallel to y-axis is given by $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2},a<b$ with the coordinates of vertices $\left(h,k\pm a\right)$ and the coordinates of foci $\left(h,k\pm c\right),c=\sqrt{a^2-b^2}$ .

Calculation:

The given information is that the ellipse has $vertices=\left(0,\pm 8\right),foci=\left(0,\pm 4\right)$, so we have the ellipse with major axis parallel to x-axis. Thus, we get

  $\begin{array}{l}\left(h,k\pm a\right)=\left(0,\pm 8\right)\\ \Rightarrow \left(h,k\right)=\left(0,0\right),a=8\end{array}$ .

Also,

  $\begin{array}{l}\left(h,k\pm c\right)=\left(0,\pm 4\right)\\ \Rightarrow c=4\\ \text{Thus, }\\ c=\sqrt{a^2-b^2}\\ 4=\sqrt{{\left(8\right)}^2-b^2}\end{array}$

Squaring both the sides, we get

  $\begin{array}{l}{\left(4\right)}^2=64-b^2\\ 16=64-b^2\\ b^2=64-16\\ b^2=48\\ b=4\sqrt{3}\end{array}$

So, the standard form of the equation is

  $\frac{x^2}{48}+\frac{y^2}{64}=1$

Conclusion:

The equation of ellipse is $\frac{x^2}{48}+\frac{y^2}{64}=1$ .