JEE Main 2014 (Online) 9th AprilQuestion 1
If $OB$ is the semi-minor axis of an ellipse, $F_1$ and $F_2$ are its foci and the angle between $F_1B$ and $F_2B$ is a right angle, then the square of the eccentricity of the ellipse is
Ellipse
Solution
Formulas used. (1) $b^2=a^2(1-e^2)$Step 1 - Right angle at $B$. $F_1(-ae,0)$, $F_2(ae,0)$, $B(0,b)$. $\vec{BF_1}\cdot\vec{BF_2}=(-ae)(ae)+(-b)(-b)=b^2-a^2e^2=0\Rightarrow b=ae$. Step 2 - Use (1). $a^2(1-e^2)=a^2e^2\Rightarrow1-e^2=e^2\Rightarrow e^2=\frac12$. Answer: C