Chapter 11.2, Problem 42E

To determine

To determine the equation of the ellipse.

Answer to Problem 42E

The equation of the ellipse is $\frac{{\text{x}}^2}{320}+\frac{{\text{y}}^2}{324}=1$ .

Explanation of Solution

Given information:

The given conditions are:

Eccentricity: $\frac{1}{9}$

Foci: $\left(0,\pm 2\right)$

Formula used:

The following formula is used:

  $\begin{array}{l}\text{e}=\frac{\text{c}}{\text{a}}\\ {\text{c}}^{\text{2}}={\text{a}}^{\text{2}}-{\text{b}}^{\text{2}}\end{array}$

Calculation:

The foci $\left(0,\pm 2\right)$ lies on the y-axis, so the ellipse has a major vertical axis and ${\text{a}}^2$ is the denominator of ${\text{y}}^2$ .

The value of $\text{a}$ can be calculated by:

  $\begin{array}{l}\frac{1}{9}=\frac{2}{\text{a}}\left[\text{e}=\frac{\text{c}}{\text{a}}\right]\\ \text{a}=18\left[\text{cross multiply}\right]\end{array}$

The value of $\text{b}$ can be calculated by:

  $\begin{array}{l}{\text{c}}^{\text{2}}={\text{a}}^{\text{2}}-{\text{b}}^{\text{2}}\\ {\text{b}}^{\text{2}}={\text{a}}^{\text{2}}-{\text{c}}^{\text{2}}\\ {\text{b}}^{\text{2}}={18}^2-2^2\\ {\text{b}}^{\text{2}}=324-4\\ {\text{b}}^{\text{2}}=300\\ \text{b}=\pm \sqrt{300}\end{array}$

The equation can be obtained by:

  $\begin{array}{l}\frac{{\text{x}}^2}{{\text{b}}^2}+\frac{{\text{y}}^2}{{\text{a}}^2}=1\\ \text{or }\frac{{\text{x}}^2}{{\sqrt{300}}^2}+\frac{{\text{y}}^2}{{18}^2}=1\\ \text{or }\frac{{\text{x}}^2}{320}+\frac{{\text{y}}^2}{324}=1\end{array}$

Hence, the equation of the ellipse is $\frac{{\text{x}}^2}{320}+\frac{{\text{y}}^2}{324}=1$ .