If the latus rectum of an ellipse is equal to half of minor axis, then its eccentricity is
If the latus rectum of an ellipse is equal to half of minor axis, then its eccentricity is
$\frac{\sqrt{3}}{4}$
$\frac{3}{4}$
$\frac{1}{4}$
$\frac{\sqrt{3}}{2}$
Solution
Let the equation of ellipse be
$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$
$\therefore$ Length of Latus rectum $=\frac{2 \mathrm{~b}^2}{\mathrm{a}}$
and length of minor axis $=2 b$
$
\begin{aligned}
& \therefore \frac{2 \mathrm{~b}^2}{\mathrm{a}}=\mathrm{b} \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{4} \\
& \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}
\end{aligned}
$