Question

If the minor axis of an ellipse subtends an angle of ${60}^{\circ }$ at each focus of the ellipse, then $4e^2=$
(where $e$ is eccentricity)

Answer 1

Clearly $△POA\cong △POB$
$\therefore \angle APO=\frac{1}{2}\angle APB=\frac{{60}^{\circ }}{2}=30°$
Now $\tan 30°=\frac{OA}{OP}\Rightarrow \frac{1}{\sqrt{3}}=\frac{b}{ae}$
$\Rightarrow e^2=\frac{3b^2}{a^2}$
$\Rightarrow e^2=3(1-e^2)[\because b^2=a^2(1-e^2\left)\right]$
$\Rightarrow 4e^2=3$