Question

If the length of latus rectum of a horizontal ellipse $x^2{\tan }^2\alpha +y^2{\sec }^2\alpha =1(\alpha \ne \frac{\pi }{2})$ is $\frac{1}{2}$, then the possible value(s) of $\alpha$ in $(0,\pi )$ is/are

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{6}$
• C. $\frac{5\pi }{12}$
• D. $\frac{7\pi }{12}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{\pi }{12}$
C $\frac{5\pi }{12}$

$x^2{\tan }^2\alpha +y^2{\sec }^2\alpha =1$
$\Rightarrow \frac{x^2}{{\cot }^2\alpha }+\frac{y^2}{{\cos }^2\alpha }=1$
$\because {\cos }^2\alpha ={\cot }^2\alpha (1-e^2)$
$\Rightarrow {\sin }^2\alpha =(1-e^2)$
$\Rightarrow e^2={\cos }^2\alpha (\alpha \ne {90}^{\circ })$
$\Rightarrow e=\cos \alpha$

$\because$ Latus rectum $=\frac{1}{2}=\frac{2b^2}{a}$
$\Rightarrow \cot \alpha =4{\cos }^2\alpha$
$\Rightarrow \frac{1}{\sin \alpha }=4\cos \alpha$
$\Rightarrow \sin 2\alpha =\frac{1}{2}$
$\Rightarrow 2\alpha =n\pi +(-1{)}^n\frac{\pi }{6}$
$\Rightarrow \alpha =\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{12}$
Since, we have to take $\alpha \in (0,\pi )$
For $n=0$
$\Rightarrow \alpha =\frac{\pi }{12}$
and for $n=1$
$\Rightarrow \alpha =\frac{\pi }{2}-\frac{\pi }{12}=\frac{5\pi }{12}$