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If the latusrectum of a hyperbola subtends a right angle at the other focus, then its eccentricity is
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Verified Answer
The correct answer is:
$\sqrt{2}+1$
$\tan \frac{\pi}{4}=\frac{L S}{S_{1} S}$
$\begin{aligned}
&=\frac{a\left(e^{2}-1\right)}{2 a e} \\
\Rightarrow \quad 1 &=\frac{e^{2}-1}{2 e} \\
\Rightarrow e^{2}-2 e-1 &=0
\end{aligned}$
Solving, $\quad e=\sqrt{2}+1$
$\begin{aligned}
&=\frac{a\left(e^{2}-1\right)}{2 a e} \\
\Rightarrow \quad 1 &=\frac{e^{2}-1}{2 e} \\
\Rightarrow e^{2}-2 e-1 &=0
\end{aligned}$
Solving, $\quad e=\sqrt{2}+1$
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