An ellipse has OB as semi-minor axis,S and S^ | vedclass

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The coordinates of the foci are $S = (ae, 0)$ and $S^{\pr

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

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The coordinates of the foci are $S = (ae, 0)$ and $S^{\prime} = (-ae, 0)$.The coordinate of the end of the semi-minor axis $B$ is $(0, b)$.Since $\angle SBS^{\prime} = 90^{\circ}$,the product of the slopes of $SB$ and $S^{\prime}B$ is $-1$.Slope of $SB = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$.Slope of $S^{\prime}B = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$.Therefore,$(-\frac{b}{ae}) 	imes (\frac{b}{ae}) = -1$.$\frac{b^2}{a^2e^2} = 1 \implies b^2 = a^2e^2$.Using the relation $b^2 = a^2(1 - e^2)$,we get $a^2e^2 = a^2(1 - e^2)$.$e^2 = 1 - e^2 \implies 2e^2 = 1 \implies e^2 = \frac{1}{2}$.Thus,$e = \frac{1}{\sqrt{2}}$.

An ellipse has $OB$ as semi-minor axis,$S$ and $S^{\prime}$ are foci and angle $\angle SBS^{\prime}$ is a right angle. Then the eccentricity of the ellipse is

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