The angle between the lines joining the foci | vedclass

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(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci are $S(ae, 0)$ and $S'(-ae, 0)$,and the extremity of the minor axis is $B(0,

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

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(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci are $S(ae, 0)$ and $S'(-ae, 0)$,and the extremity of the minor axis is $B(0, b)$.Given that the angle $\angle SBS' = 90^{\circ}$.Since the triangle $	riangle SBS'$ is isosceles with $SB = S'B$,the altitude from $B$ to $SS'$ bisects the angle $\angle SBS'$.Thus,$\angle OBS = 45^{\circ}$.In the right-angled triangle $	riangle OBS$,we have $	an(45^{\circ}) = \frac{OB}{OS} = \frac{b}{ae}$.Since $	an(45^{\circ}) = 1$,we get $1 = \frac{b}{ae}$,which implies $b = ae$.Using the relation $b^2 = a^2(1 - e^2)$,we substitute $b^2 = a^2e^2$:$a^2e^2 = a^2(1 - e^2)$$e^2 = 1 - e^2$$2e^2 = 1$$e^2 = \frac{1}{2}$$e = \frac{1}{\sqrt{2}}$.

The angle between the lines joining the foci of an ellipse to one particular extremity of the minor axis is $90^{\circ}$. The eccentricity of the ellipse is

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