Chapter 7, Problem 1MCCP

In Exercises 1-5, graph each ellipse. Give the location of the foci.

$\frac{x^2}{25}+\frac{y^2}{4}=1$

To determine

The graph of the ellipse $\frac{x^2}{25}+\frac{y^2}{4}=1$ and location of foci.

Answer to Problem 1MCCP

Solution: The foci are at $\left(\pm \sqrt{21},0\right)$ and graph is

Explanation of Solution

Given: $\frac{x^2}{25}+\frac{y^2}{4}=1$

Find the values of $a$ and $b$ from $\frac{x^2}{25}+\frac{y^2}{4}=1$ by comparing it with the conventional equation of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as

$\begin{array}{c}{a}^2=25\\ a=5\end{array}$

and

$\begin{array}{c}{b}^2=25\\ b=2\end{array}$

The focus of the ellipse lies at $\left(\pm ae,0\right)$:

$\begin{array}{c}e=\sqrt{1-\frac{b^2}{a^2}}\\ =\sqrt{1-\frac{4}{25}}\\ =\sqrt{\frac{21}{25}}\\ =\frac{\sqrt{21}}{5}\end{array}$

$\begin{array}{c}ae=5\cdot \left(\frac{\sqrt{21}}{5}\right)\\ =\sqrt{21}\end{array}$

Therefore, the foci lies at $\left(\pm \sqrt{21},0\right)$.

The graph of the ellipse is plotted as

Conclusion: The graph is plotted and the foci are at $\left(\pm \sqrt{21},0\right)$.