Q93 / 160 · AP EAMCET 2022 (06 Jul Shift 1) · Mathematics
If the latus rectum of an ellipse is equal to half of minor axis, then its eccentricity is
- A$\frac{\sqrt{3}}{4}$
- B$\frac{3}{4}$
- C$\frac{1}{4}$
- Correct answerD$\frac{\sqrt{3}}{2}$
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 $=2 b$
$$
\begin{aligned}
& \therefore \frac{2 \mathrm{~b}^2}{\mathrm{a}}=\mathrm{b} \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{4} \\
& \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.