If P(frac{pi}{6}) is a point on the hyperbola | vedclass

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(C) For a point $P$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the focal distances are $SP = |ex - a|$ and $S^{\prime}P = |ex + a|$.Give

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$\sqrt{3}$

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(C) For a point $P$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the focal distances are $SP = |ex - a|$ and $S^{\prime}P = |ex + a|$.Given the condition $SP + S^{\prime}P - 2|SP - S^{\prime}P| = 0$,we know that $|SP - S^{\prime}P| = 2a$.Substituting this into the equation: $SP + S^{\prime}P = 2(2a) = 4a$.For a hyperbola,the sum of focal distances is $SP + S^{\prime}P = 2ex$ (assuming $x > 0$).Thus,$2ex = 4a$,which implies $ex = 2a$,or $x = \frac{2a}{e}$.The point $P$ is given as $P(\frac{\pi}{6})$,which in parametric form is $(a \sec 	heta, b 	an 	heta)$.Here,$x = a \sec(\frac{\pi}{6}) = a \cdot \frac{2}{\sqrt{3}}$.Equating the two expressions for $x$: $\frac{2a}{e} = \frac{2a}{\sqrt{3}}$.Therefore,$e = \sqrt{3}$.

If $P(\frac{\pi}{6})$ is a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,$S$ and $S^{\prime}$ are its foci,and $SP + S^{\prime}P - 2|SP - S^{\prime}P| = 0$,then the eccentricity $e$ is:

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