If S and S^{prime} are the foci of an ellipse | vedclass

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(C) 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^{\prim

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

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(C) 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 one end of the minor axis $B$ is $(0, b)$.The slope of $SB$ is $m_1 = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$.The slope of $S^{\prime}B$ is $m_2 = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$.Given $\angle SBS^{\prime} = 90^{\circ}$,the product of the slopes is $-1$:$m_1 	imes m_2 = -1$$\left(\frac{b}{ae}ight) 	imes \left(-\frac{b}{ae}ight) = -1$$\frac{b^2}{a^2e^2} = 1 \Rightarrow b^2 = a^2e^2$.We know that for an ellipse,$b^2 = a^2(1 - e^2)$.Substituting $b^2 = a^2e^2$ into the relation:$a^2e^2 = a^2(1 - e^2)$$e^2 = 1 - e^2$$2e^2 = 1$$e^2 = \frac{1}{2} \Rightarrow e = \frac{1}{\sqrt{2}}$.

If $S$ and $S^{\prime}$ are the foci of an ellipse,$B$ is one end of the minor axis and $\angle S B S^{\prime} = 90^{\circ}$,then the eccentricity of that ellipse is

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