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)

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