Question

Find the equation of the ellipse whose foci are $(4,0)$ and $(-4,0),$ eccentricity $e=\frac{1}{3}$.

Answer 1

Given: Foci are $(4,0)$ and $(-4,0),$ eccentricity $e=\frac{1}{3}$

This is an ellipse of form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$

Foci $=(\pm ae,0)$

$\Rightarrow ae=4$

$\Rightarrow a=12$ $[\because e=\frac{1}{3}]$

$b^2=a^2-a^2{e}^2$

$\Rightarrow b^2=144-16$

$\Rightarrow b^2=128$

So, the equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{x^2}{144}+\frac{y^2}{128}=1$

Hence, the equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{8}=16$