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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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The correct answer is:
$\frac{1}{\sqrt{2}}$
$S$ and $S^1$ are the foci of an ellipse
$\therefore$ Coordinate of foci $=( \pm a e, 0)$
Slope of $S B=\frac{b}{a e}$
Slope of $S^1 B=\frac{-b}{a e}$
Now, $\angle S B S^{\prime}=90$
So, slope of $S B \times$ slope of $S^{\prime} B=-1$
$$
\begin{array}{ll}
\Rightarrow & \frac{b}{a e} \times\left(-\frac{b}{a e}\right)=-1 \\
\Rightarrow & \frac{b^2}{a^2 e^2}=1 \Rightarrow b^2=a^2 e^2
\end{array}
$$
We know that,
$$
\begin{aligned}
b^2 & =a^2\left(1-e^2\right) \\
\Rightarrow \quad a^2 e^2 & =a^2\left(1-e^2\right) \Rightarrow e^2=1-e^2 \\
\Rightarrow \quad 2 e^2 & =1 \Rightarrow e=\frac{1}{\sqrt{2}}
\end{aligned}
$$
$\therefore$ Coordinate of foci $=( \pm a e, 0)$
Slope of $S B=\frac{b}{a e}$
Slope of $S^1 B=\frac{-b}{a e}$
Now, $\angle S B S^{\prime}=90$
So, slope of $S B \times$ slope of $S^{\prime} B=-1$
$$
\begin{array}{ll}
\Rightarrow & \frac{b}{a e} \times\left(-\frac{b}{a e}\right)=-1 \\
\Rightarrow & \frac{b^2}{a^2 e^2}=1 \Rightarrow b^2=a^2 e^2
\end{array}
$$
We know that,
$$
\begin{aligned}
b^2 & =a^2\left(1-e^2\right) \\
\Rightarrow \quad a^2 e^2 & =a^2\left(1-e^2\right) \Rightarrow e^2=1-e^2 \\
\Rightarrow \quad 2 e^2 & =1 \Rightarrow e=\frac{1}{\sqrt{2}}
\end{aligned}
$$
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