Question

If the latusrectum of an ellipse be equal to half of its minor axis, then its eccentricity is ___________.

Answer 1

For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where a > b
Major axis is given by 2a and minor axis is 2b
$e^2=\frac{a^2-b^2}{a^2}...\left(1\right)$
Since latus rectum is $\frac{2b^2}{a}...\left(2\right)$
Given latus rectum is half of its minor axis
i.e $\frac{2b^2}{a}=\frac{1}{2}\times 2b$
i.e a = 2b
$$\begin{gathered} \mathrm{i}.\mathrm{e}{e}^2=\frac{(a^2-b^2)}{a^2} \\ =\frac{4b^2-b^2}{{\left(2b\right)}^2} \\ {e}^2=\frac{3b^2}{4b^2} \\ \mathrm{i}.\mathrm{e}e=\frac{\sqrt{3}}{2} \end{gathered}$$