Q. $B$ is an extremity of the minor axis of an ellipse whose foci are $S$ and $S'$. If $∠SBS'$ is a right angle, then the eccentricity of the ellipse is

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$\frac{1}{2}$

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

$\frac{2}{3}$

$\frac{1}{3}$

Solution:

Slope of $SB, m_1 = \frac{b - 0}{0 - ae} =-\frac{b}{ae}$
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 \frac{-b}{a e} \times \frac{b}{a e} =-1 $
$\Rightarrow b^{2} =a^{2} e^{2} $
$\Rightarrow \frac{b^{2}}{a^{2}} =e^{2} $
$\Rightarrow 1-e^{2} =e^{2} $
$\Rightarrow 2 e^{2} =1$
$\Rightarrow e^{2}=\frac{1}{2} $
$\Rightarrow e=\pm \frac{1}{\sqrt{2}} $
$ \Rightarrow e=\frac{1}{\sqrt{2}} $
or $ e=-\frac{1}{\sqrt{2}}$

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