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Let $S$ and $S^{\prime}$ be the foci of an ellipse and $B$ be one end of its minor axis. If $S B S^{\prime}$ is a isosceles right angled triangle then the eccentricity of the ellipse is
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{2}}$
We have,
$S B S^{\prime}$ is an isosceles right angle triangle.
$$
\begin{aligned}
& \therefore \quad S S^2=S B^2+S^{\prime} B^2 \\
& \Rightarrow \quad(2 a e)^2=b^2+(a e)^2+b^2+(a e)^2 \\
& \Rightarrow \quad 4(a e)^2=2\left(b^2+(a e)^2\right) \\
& \Rightarrow \quad(a e)^2=b^2 \\
& \Rightarrow \quad e^2=\frac{b^2}{a^2} \\
& \Rightarrow \quad 1-e^2=1-\frac{b^2}{a^2} \\
& \Rightarrow \quad 1-e^2=e^2 \\
& {\left[\because e=\sqrt{1-\frac{b^2}{a^2}}\right]} \\
& \Rightarrow \quad 2 e^2=1 \\
& \Rightarrow \quad e=\frac{1}{\sqrt{2}} \\
&
\end{aligned}
$$
$S B S^{\prime}$ is an isosceles right angle triangle.
$$
\begin{aligned}
& \therefore \quad S S^2=S B^2+S^{\prime} B^2 \\
& \Rightarrow \quad(2 a e)^2=b^2+(a e)^2+b^2+(a e)^2 \\
& \Rightarrow \quad 4(a e)^2=2\left(b^2+(a e)^2\right) \\
& \Rightarrow \quad(a e)^2=b^2 \\
& \Rightarrow \quad e^2=\frac{b^2}{a^2} \\
& \Rightarrow \quad 1-e^2=1-\frac{b^2}{a^2} \\
& \Rightarrow \quad 1-e^2=e^2 \\
& {\left[\because e=\sqrt{1-\frac{b^2}{a^2}}\right]} \\
& \Rightarrow \quad 2 e^2=1 \\
& \Rightarrow \quad e=\frac{1}{\sqrt{2}} \\
&
\end{aligned}
$$
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