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Let $\mathrm{A}, \mathrm{A}^{\prime}$ be the end points of major axis, $\mathrm{S}, \mathrm{S}^{\prime}$ be the foci and $\mathrm{B}, \mathrm{B}^{\prime}$ be the end points of minor axis of an ellipse $\mathrm{E}$. If $\angle \mathrm{BAB}^{\prime}=60^{\circ}$, then $\angle \mathrm{SBS}^{\prime}=$
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The correct answer is:
$\tan ^{-1}(-2 \sqrt{2})$
$\tan 30^{\circ}=\frac{b}{a} \Rightarrow a=\sqrt{3} b$
We have $c=\sqrt{a^2-b^2}= \pm \sqrt{2} b$ Now, $\tan \theta=\frac{c}{b}=\frac{ \pm \sqrt{2} b}{b}= \pm \sqrt{2}$,
$\begin{aligned} & \angle \mathrm{SBS}^{\prime}=2 \theta=2 \tan ^{-1}( \pm \sqrt{2}) \\ & =\tan ^{-1}\left(\frac{2 \sqrt{2}}{1-2}\right)=\tan ^{-1}(-2 \sqrt{2})\end{aligned}$
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