Q31 / 90 · JEE Main 2014 (09 Apr Online) · Mathematics

If $O B$ is the semi-minor axis of an ellipse, $F_{1}$ and $F_{2}$ are its focii and the angle between $F_{1} B$ and $F_{2} B$ is a right angle, then the square of the eccentricity of the ellipse is

Solution

In standard ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Semi-major axis $O A = a$

Semi-minor axis $O B = b$

Focus $F_{1} \equiv (- a e, 0)$

$F_{2} \equiv (a e, 0)$

$∠ F_{1} B F_{2} = 90 °$

$O F_{1} = O F_{2} \Rightarrow ∠ F_{1} B O = ∠ F_{2} B O = 45 °$

$\Rightarrow \frac{a e}{b} = tan 45 ° = 1$

$\Rightarrow b = a e$

Also, $b^{2} = a^{2} (1 - e^{2})$

$\Rightarrow a^{2} e^{2} = a^{2} (1 - e^{2})$

$\Rightarrow 2 e^{2} = 1$

$\Rightarrow e^{2} = \frac{1}{2}$

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