Consider an ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 .$ Points P and Q are two points on ellipse and A and B are the corresponding points of P and Q on the auxiliary circle of ellipse. O is origin, then find the correct options

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If OA is perpendicular to OB and Q’ is reflection of Q is origin then minimum value of POQ’ is $\tan^{- 1} (\frac{12}{5})$

The maximum value of angle AOP is $\tan^{- 1} (2 \sqrt{2})$

If OA is perpendicular to OB and Q’ is reflection of Q in origin then minimum value of angle POQ’ is $\tan^{- 1} (\frac{4}{3})$

The maximum value of angle AOP is $\tan^{- 1} (\frac{1}{2 \sqrt{6}})$

answer is B, D.

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Detailed Solution

$\begin{aligned} \tan (AOP) = \frac{\tan θ - \frac{b}{a} \tan θ}{1 + \frac{b}{a} \tan^{2} θ} \\ = \frac{1 - \frac{b}{a}}{\cot θ + \frac{b}{a} \tan θ} \leq \frac{a - b}{2 \sqrt{ab}} \\ max (AOP) = \tan^{- 1} (\frac{a - b}{2 \sqrt{ab}}) \end{aligned} \begin{array}{l} = \tan^{- 1} (\frac{3 - 2}{2 \sqrt{6}}) = \tan^{- 1} (\frac{1}{2 \sqrt{6}}) \\ \tan ({POQ}^{'}) = \frac{\frac{b}{a} \tan θ + \frac{b}{a} \cot θ}{1 - \frac{b^{2}}{a^{2}}} \geq \frac{a}{\frac{a^{2} - b^{2}}{a^{2}}} \geq \frac{2 ab}{a^{2} - b^{2}} \end{array}$
So, min angle POQ’ is $\tan^{- 1} (\frac{2 \times 3 \times 2}{9 - 4})$
$= \tan^{- 1} (\frac{12}{5})$