An ellipse has OB as semi-minor axis, F and F’ its foci and the angle FBF’ is a right angle. Then the eccentricity of the ellipse is

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Solution

The correct option is A
$tan 45^{\circ} = \frac{a e}{b} \Rightarrow a e = b$
$\Rightarrow a^{2} e^{2} = b^{2} = a^{2} (1 - e^{2})$
$\Rightarrow e = \frac{1}{\sqrt{2}}$

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