Question

Let A, A' be the end points of major axis, S, S' be the foci and B, B' be the end points of minor axis of an ellipse E. If $\angle \text{BAB}' = 60^\circ$, then $\angle \text{SBS}' =$

• A. $\tan^{-1}(\sqrt{2})$
• B. $\tan^{-1}(-2\sqrt{2})$
• C. $\tan^{-1}\left(\sqrt{\frac{2}{3}}\right)$
• D. $\tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$

Hint:

Ellipse Essentials:

• $b^2 = a^2(1 - e^2)$
• Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$
• $\cos\alpha = x \Rightarrow \tan\alpha = \pm\sqrt1 - x^2/x$
• Use triangle symmetry and angle properties to relate geometry with algebra.

Answer 1

The Correct Option is B

Assume ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a>b$. Points: $A=(a,0)$, $A'=(-a,0)$, $B=(0,b)$, $B'=(0,-b)$, $S=(ae,0)$, $S'=(-ae,0)$, $e^2 = 1 - \frac{b^2}{a^2}$ Given: $\angle BAB' = 60^\circ$, triangle $BAB'$ is equilateral: \[ AB = BB' \Rightarrow \sqrt{a^2 + b^2} = 2b \Rightarrow a^2 = 3b^2 \] Then: \[ b^2 = a^2(1 - e^2) \Rightarrow \frac{a^2}{3} = a^2(1 - e^2) \Rightarrow e^2 = \frac{2}{3} \] In triangle $SBS'$, $S=(ae,0)$, $B=(0,b)$, $S'=(-ae,0)$ \[ SS' = 2ae, \quad SB = S'B = a \] Using cosine rule: \[ % Option (2ae)^2 = 2a^2 - 2a^2\cos\alpha \Rightarrow \cos\alpha = 1 - 2e^2 = -\frac{1}{3} \] \[ \tan^2\alpha = \frac{1}{\cos^2\alpha} - 1 = 9 - 1 = 8 \Rightarrow \tan\alpha = -2\sqrt{2} \]