Question

If the eccentricity of an ellipse is $\frac{5}{8}$ and the distance between its foci is 10, then find the latus rectum of the ellipse.

Answer 1

Let the equation of the required ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and let e be its eccentricity.

We have, $e=\frac{5}{8}$ and $2ae=10$

$e=\frac{5}{8}$ and $ae=5$

$e=\frac{5}{8}$ and $a=8$

Therefore, $b^2=a^2(1-e^2)$

$b^2=64(1-\frac{25}{64})$

$b^2=39$

Hence, length of the latusrectum = $\frac{2b^2}{a}=2\times \frac{39}{8}=\frac{39}{4}$