Let $S$ and $S^{\prime}$ be the foci of an ellipse and $B$ be one end of its minor axis. If $S B S^{\prime}$ is a isosceles right angled triangle then the eccentricity of the ellipse is

• Correct answer
  A

  $\frac{1}{\sqrt{2}}$
• B

  $\frac{1}{2}$
• C

  $\frac{\sqrt{3}}{2}$
• D

  $\frac{1}{3}$

Solution

We have, $S B S^{\prime}$ is an isosceles right angle triangle.
$$ \begin{aligned} & \therefore \quad S S^2=S B^2+S^{\prime} B^2 \\ & \Rightarrow \quad(2 a e)^2=b^2+(a e)^2+b^2+(a e)^2 \\ & \Rightarrow \quad 4(a e)^2=2\left(b^2+(a e)^2\right) \\ & \Rightarrow \quad(a e)^2=b^2 \\ & \Rightarrow \quad e^2=\frac{b^2}{a^2} \\ & \Rightarrow \quad 1-e^2=1-\frac{b^2}{a^2} \\ & \Rightarrow \quad 1-e^2=e^2 \\ & {\left[\because e=\sqrt{1-\frac{b^2}{a^2}}\right]} \\ & \Rightarrow \quad 2 e^2=1 \\ & \Rightarrow \quad e=\frac{1}{\sqrt{2}} \\ & \end{aligned} $$

Hence, the correct answer is option: 'A'.