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: \[ % Option (2ae)^2 = 2a^2 - 2a^2\cos\alpha \Rightarrow \cos\alpha = 1 - 2e^2 = -\frac{1}{3} \] \[ \tan^2\alpha = \frac{1}{\cos^2\alpha} - 1 = 9 - 1 = 8 \Rightarrow \tan\alpha = -2\sqrt{2} \]

Answer 2

Step 1: Understand the ellipse parameters
Let the ellipse be centered at the origin with standard form:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where \(a > b > 0\).

Step 2: Identify points on ellipse
- \(A, A'\) are endpoints of the major axis along the x-axis: \(A = (a, 0)\), \(A' = (-a, 0)\)
- \(B, B'\) are endpoints of the minor axis along the y-axis: \(B = (0, b)\), \(B' = (0, -b)\)
- \(S, S'\) are foci: \(S = (c, 0)\), \(S' = (-c, 0)\), where \(c^2 = a^2 - b^2\)

Step 3: Use given angle \(\angle BAB' = 60^\circ\)
This angle is formed at \(A\) by points \(B\) and \(B'\).
Vectors from \(A\) are:
\[ \overrightarrow{AB} = (0 - a, b - 0) = (-a, b), \quad \overrightarrow{AB'} = (0 - a, -b - 0) = (-a, -b) \]
The angle between these vectors is 60°, so:
\[ \cos 60^\circ = \frac{\overrightarrow{AB} \cdot \overrightarrow{AB'}}{|\overrightarrow{AB}| |\overrightarrow{AB'}|} \]
Calculate dot product:
\[ (-a)(-a) + b(-b) = a^2 - b^2 \]
Calculate magnitudes:
\[ |\overrightarrow{AB}| = \sqrt{(-a)^2 + b^2} = \sqrt{a^2 + b^2} \] Similarly, \(|\overrightarrow{AB'}| = \sqrt{a^2 + b^2}\)
Thus,
\[ \cos 60^\circ = \frac{a^2 - b^2}{a^2 + b^2} = \frac{c^2}{a^2 + b^2} \]
Since \(\cos 60^\circ = \frac{1}{2}\), we get:
\[ \frac{c^2}{a^2 + b^2} = \frac{1}{2} \implies 2c^2 = a^2 + b^2 \]

Step 4: Use relation \(c^2 = a^2 - b^2\)
Substitute \(c^2\) in above:
\[ 2(a^2 - b^2) = a^2 + b^2 \implies 2a^2 - 2b^2 = a^2 + b^2 \]\[ 2a^2 - a^2 = 2b^2 + b^2 \implies a^2 = 3b^2 \]

Step 5: Calculate \(\angle SBS'\)
At point \(B\), vectors:
\[ \overrightarrow{BS} = (c - 0, 0 - b) = (c, -b), \quad \overrightarrow{BS'} = (-c - 0, 0 - b) = (-c, -b) \]
Angle between these vectors is:
\[ \cos \theta = \frac{\overrightarrow{BS} \cdot \overrightarrow{BS'}}{|\overrightarrow{BS}| |\overrightarrow{BS'}|} \]
Dot product:
\[ c \times (-c) + (-b) \times (-b) = -c^2 + b^2 = b^2 - c^2 \]
Magnitudes:
\[ |\overrightarrow{BS}| = \sqrt{c^2 + b^2}, \quad |\overrightarrow{BS'}| = \sqrt{c^2 + b^2} \]
So,
\[ \cos \theta = \frac{b^2 - c^2}{c^2 + b^2} \]
Substitute \(c^2 = a^2 - b^2\):
\[ \cos \theta = \frac{b^2 - (a^2 - b^2)}{(a^2 - b^2) + b^2} = \frac{b^2 - a^2 + b^2}{a^2} = \frac{2b^2 - a^2}{a^2} \]
Using \(a^2 = 3b^2\), we get:
\[ \cos \theta = \frac{2b^2 - 3b^2}{3b^2} = \frac{-b^2}{3b^2} = -\frac{1}{3} \]

Step 6: Find \(\tan \theta\)
\[ \cos \theta = -\frac{1}{3} \implies \theta = \cos^{-1}\left(-\frac{1}{3}\right) \]
Calculate \(\tan \theta\) using identity:
\[ \tan \theta = \pm \sqrt{\frac{1}{\cos^2 \theta} - 1} = \pm \sqrt{\frac{1}{\frac{1}{9}} - 1} = \pm \sqrt{9 - 1} = \pm \sqrt{8} = \pm 2\sqrt{2} \]
Since \(\cos \theta\) is negative and \(\theta\) is in second quadrant, \(\tan \theta < 0\), so:
\[ \tan \theta = -2\sqrt{2} \]

Conclusion:
\[ \angle SBS' = \tan^{-1}(-2\sqrt{2}) \]