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If $\mathrm{P}\left(\frac{\pi}{6}\right)$ is a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \mathrm{~S}, \mathrm{~S}^{\prime}$ are its foci and $\mathrm{SP}+\mathrm{S}^{\prime} \mathrm{P}-2\left|\mathrm{SP}-\mathrm{S}^{\prime} \mathrm{P}\right|$, then $\mathrm{e}=$
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Verified Answer
The correct answer is:
$\sqrt{3}$
$$
\begin{aligned}
& \text { } \because\left|P S-P S^{\prime}\right|=2 a \\
& \Rightarrow 2|P S-P S|=4 a \\
& \text { now } P S+P S=2 e \cdot \frac{2 a}{\sqrt{3}}
\end{aligned}
$$
$\left(\right.$ where $\frac{2 \mathrm{a}}{\sqrt{3}}$ is the $\mathrm{x}$ - coordinate of point $\left.\mathrm{P}\right)$
$$
\Rightarrow \mathrm{e}=\sqrt{3}
$$
\begin{aligned}
& \text { } \because\left|P S-P S^{\prime}\right|=2 a \\
& \Rightarrow 2|P S-P S|=4 a \\
& \text { now } P S+P S=2 e \cdot \frac{2 a}{\sqrt{3}}
\end{aligned}
$$
$\left(\right.$ where $\frac{2 \mathrm{a}}{\sqrt{3}}$ is the $\mathrm{x}$ - coordinate of point $\left.\mathrm{P}\right)$
$$
\Rightarrow \mathrm{e}=\sqrt{3}
$$
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