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(B) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.Let the foci be $F(ae, 0)$ and $F'(-ae, 0)$ and the end point of the mino

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$1/\sqrt{2}$

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(B) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.Let the foci be $F(ae, 0)$ and $F'(-ae, 0)$ and the end point of the minor axis be $B(0, b)$.The center of the ellipse is $C(0, 0)$.Given that $\angle FBF' = \frac{\pi}{2}$.Since $	riangle FBF'$ is an isosceles triangle with $BF = BF'$,the altitude $BC$ bisects $\angle FBF'$.Therefore,$\angle FBC = \frac{1}{2} \angle FBF' = \frac{\pi}{4}$.In the right-angled triangle $	riangle BCF$,$	an(\angle FBC) = \frac{CF}{BC}$.$	an(\frac{\pi}{4}) = \frac{ae}{b}$.$1 = \frac{ae}{b} \Rightarrow b = ae$.Squaring both sides,$b^2 = a^2 e^2$.Using the relation $b^2 = a^2(1 - e^2)$,we get $a^2(1 - e^2) = a^2 e^2$.$1 - e^2 = e^2$.$2e^2 = 1$.$e^2 = 1/2$.$e = 1/\sqrt{2}$.

If the angle between the lines joining the end points of the minor axis of an ellipse with its foci is $\frac{\pi}{2}$,then the eccentricity of the ellipse is

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