If the angle between the lines joining the end points of minor axis of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with one of its foci is $\frac{\pi}{2}$, then what is the eccentricity of the ellipse?

• A

  $\frac{1}{2}$
• Correct answer
  B

  $\frac{1}{\sqrt{2}}$
• C

  $\frac{\sqrt{3}}{2}$
• D

  $\frac{1}{2 \sqrt{2}}$

Solution

Given, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
given: $\angle P F_1 Q=90^{\circ}$ $\begin{array}{ll} \therefore & m_{P F_1} \times m_{Q F_1}=-1 \\ \Rightarrow & \frac{-b}{a e} \times \frac{b}{a e}=-1 \\ \Rightarrow & b^2=a^2 \cdot e^2 \cdots(i)\\ \because & b^2=a^2\left(1-e^2\right)...(ii) \end{array}$ From (i) and (ii) we get, $\begin{aligned} & e^2=1-e^2 \\ & \Rightarrow 2 e^2=1 \\ & \Rightarrow e^2=\frac{1}{2} \\ & \Rightarrow e=\frac{1}{\sqrt{2}} \end{aligned}$

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