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$B$ is an extremity of the minor axis of an ellipse whose foci are $S$ and $S^{\prime} .$ If $\angle S B S^{\prime}$ is a right angle, then the eccentricity of the ellipse is
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The correct answer is:
$\frac{1}{\sqrt{2}}$
Slope of $S B, m_{1}=\frac{b-0}{0-a e}=-\frac{b}{a e}$
and slope of $S^{\prime} B, m_{2}=\frac{b-0}{0-(-a e)}=\frac{b}{a e}$
since, $\angle S B S^{\prime}$ is a right angle.
$\therefore$ $$m_{1} m_{2}=-1$$
$\Rightarrow \quad \frac{-b}{a e} \times \frac{b}{a e}=-1$
$\Rightarrow \quad b^{2}=a^{2} e^{2}$
$\Rightarrow \quad \frac{b^{2}}{a^{2}}=e^{2}$
$\Rightarrow \quad 1-e^{2}=e^{2}$
$\Rightarrow \quad 2 e^{2}=1$
$\Rightarrow \quad e^{2}=\frac{1}{2}$
$\Rightarrow \quad e=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \quad e=\frac{1}{\sqrt{2}}$ or $e=-\frac{1}{\sqrt{2}}$
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