Question

An ellipse has OB as semi-minor axis, S and S' are foci and angle SBS' is a right angle. Then the eccentricity of the ellipse is

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\sqrt{2}$
• D. $\frac{1}{3}$

Hint:

Use Pythagoras when angle between focal lines is $90^\circ$.

Answer 1

The Correct Option is B

Concept:
For ellipse: \[ c^2 = a^2 - b^2,\quad e = \frac{c}{a} \] Step 1: Use geometric condition. At point $B(0,b)$: \[ SB = S'B = \sqrt{c^2 + b^2} \] Given $\angle SBS' = 90^\circ$: \[ SS'^2 = SB^2 + S'B^2 \] \[ (2c)^2 = 2(c^2 + b^2) \] \[ 4c^2 = 2c^2 + 2b^2 \Rightarrow 2c^2 = 2b^2 \Rightarrow c^2 = b^2 \]
Step 2: Substitute. \[ c^2 = a^2 - b^2 \Rightarrow b^2 = a^2 - b^2 \] \[ 2b^2 = a^2 \Rightarrow b^2 = \frac{a^2}{2} \] \[ c^2 = \frac{a^2}{2} \]
Step 3: Find eccentricity. \[ e = \frac{c}{a} = \frac{1}{\sqrt{2}} \]
Step 4: Conclusion. \[ {\frac{1}{\sqrt{2}}} \]