Question

Points $A$ and $B$ lie on the auxiliary circle of ellipse $\frac{x^4}{4}+y^2=1.$ $P$ and $Q$ are the corresponding points on the ellipse for the points $A$ and $B$ respectively ($O$ is the origin). Which of the following options is/are CORRECT?

• A. The maximum value of angle $AOP$ is ${\tan }^{-1}\left(2\sqrt{2}\right)$
• B. The maximum value of angle $AOP$ is ${\tan }^{-1}\left(\frac{1}{2\sqrt{2}}\right)$
• C. If $OA\perp OB$ and $Q^{\prime }$ is reflection of $Q$ in origin, then the minimum value of angle $PO{Q}^{\prime }$ is ${\tan }^{-1}\left(\frac{4}{3}\right)$
• D. If $OA\perp OB$ and $Q^{\prime }$ is reflection of $Q$ in origin, then the minimum value of angle $PO{Q}^{\prime }$ is ${\tan }^{-1}\left(\frac{3}{4}\right)$

Answer 1

Answer: (B), (C)

If $OA\perp OB$ and $Q^{\prime }$ is reflection of $Q$ in origin, then the minimum value of angle $PO{Q}^{\prime }$ is ${\tan }^{-1}\left(\frac{4}{3}\right)$

Let $\angle AOP=\alpha$
Using formula of angle between two points,
$\tan \alpha =\frac{\tan \theta -\frac{b}{a}\tan \theta }{1+\frac{b}{a}{\tan }^2\theta }$
$=\frac{1-\frac{b}{a}}{\cot \theta +\frac{b}{a}\tan }\le \frac{1-\frac{b}{a}}{2\sqrt{\frac{b}{a}}}=\frac{a-b}{2\sqrt{ab}}$

$a=2$ and $b=1$ (Given)
$\tan \alpha \le \frac{a-b}{2\sqrt{ab}}=\frac{2-1}{2\sqrt{2\times 1}}=\frac{1}{2\sqrt{2}}$

For $\angle PO{Q}^{\prime }$
$B(a\cos (\frac{\pi }{2}+\theta ),a\sin (\frac{\pi }{2}+\theta ))\equiv B(-a\sin \theta ,a\cos \theta )$
$Q(-a\sin \theta ,b\cos \theta ),Q^{\prime }(a\sin \theta ,-b\cos \theta )$

Let $\angle PO{Q}^{\prime }=\beta$
$\tan \beta =\frac{\frac{b}{a}\tan \theta +\frac{b}{a}\cot \theta }{1-\frac{b^2}{a^2}}\ge \frac{\frac{2b}{a}}{1-\frac{b^2}{a^2}}$
$\tan \beta \ge \frac{2ab}{a^2-b^2}=\frac{2\cdot 2\cdot 1}{4-1}=\frac{4}{3}$