Chapter 11.2, Problem 33E

To determine

To determine the equation of the ellipse.

Answer to Problem 33E

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

Explanation of Solution

Given information:

The given conditions are:

  $\text{Foci:}\left(\pm 4,0\right),\text{ Vertices:}\left(\pm 5,0\right)$

Formula used:

The following formula is used:

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

Calculation:

The foci and vertices are of the form $\left(\pm \text{x},0\right)$ ,so the ellipse has a major horizontal axis and ${\text{a}}^2$ is the denominator of ${\text{x}}^2$ .

The value of $\text{a and c are }\pm \text{5 and }\pm \text{4 respectively}$

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

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

The equation can be obtained by:

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

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