Question

If latus rectum of the ellipse $x^2 \tan ^2 \alpha+y^2 \sec ^2 \alpha=1$ is $\frac{1}{2}$, where, $0<\alpha<\pi$, then eccentricity ${ }_{-}^{\prime} e^{\prime}$ can be

• A. $\frac{\sqrt{3} + 1}{2 \sqrt{2}}$
• B. $\frac{\sqrt{3}}{2 \sqrt{2}}$
• C. $\frac{1}{2 \sqrt{2}}$
• D. none of these

Answer 1

Answer: (A)

$x^2 \tan ^2 \alpha+y^2 \sec ^2 \alpha=1$
$\Rightarrow \frac{x^2}{\cot ^2 \alpha}+\frac{y^2}{\cos ^2 \alpha}=1$
$\because(\cos )^2 \alpha=(\cot )^2 \alpha\left(1-e^2\right)$
$\Rightarrow(\sin )^2 \alpha=\left(1-e^2\right)$
$\therefore e^2=(\cos )^2 \alpha\left(\alpha \neq 90^{\circ}\right)$
$e=\cos \alpha$
$\because$ Latus rectum $=1 / 2=2 b^2 / a$
$\Rightarrow a=4 b^2$,
$\Rightarrow \cot \alpha=4 \cos ^2 \alpha$
$\Rightarrow \frac{1}{\sin \alpha}=4 \cos \alpha$
$\Rightarrow \sin 2 \alpha=\frac{1}{2}$