Chapter 11.2, Problem 40E

To determine

To determine the equation of the ellipse.

Answer to Problem 40E

The equation of the ellipse is $\frac{{\text{x}}^2}{25}+\frac{{\text{y}}^2}{9}=1$ .

Explanation of Solution

Given information:

The given conditions are:

Endpoints of the minor axis: $\left(0,\pm 3\right)$

Distance between foci: $8$

Formula used:

The following formula is used:

  ${\text{c}}^2={\text{a}}^2-{\text{b}}^2$

Calculation:

The y-axis is the minor axis because endpoints of the minor axis $\left(0,\pm 3\right)$ lies on the y-axis, and x-axis is the major axis so the ellipse has a major horizontal axis and ${\text{a}}^2$ is the denominator of ${\text{x}}^2$ .

The value of $\text{b}=\pm 3$

The distance between the foci is $8$ , so the foci: $\left(\pm 4,0\right)$

The value of $\text{c is }\pm 4$

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

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

The equation can be obtained by:

  $\begin{array}{l}\frac{{\text{x}}^2}{{\text{a}}^2}+\frac{{\text{y}}^2}{{\text{b}}^2}=1\\ \text{or }\frac{{\text{x}}^2}{5^2}+\frac{{\text{y}}^2}{3^2}=1\\ \text{or }\frac{{\text{x}}^2}{25}+\frac{{\text{y}}^2}{9}=1\end{array}$

Hence, the equation of the ellipse is $\frac{{\text{x}}^2}{25}+\frac{{\text{y}}^2}{9}=1$ .