If OB is the semi-minor axis of on ellipse, $F_{1}$ and $F_{2}$ are its foci 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

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$\frac{1}{2}$

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

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

$\frac{1}{4}$

answer is A.

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Detailed Solution

Let the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)$

where ${OF}_{1} = ae, OB = b$

from $∆^{le} {BOF}_{1}, \tan 45^{0} = \frac{ae}{b}$

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

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