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The tangent at a point P on the hyperbola $\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}$ meets one of its directrices at point F. If PF subtends an angles $\mathrm{\theta}$ at the corresponding focus, then $\mathrm{\theta}$ equals
Option: 1
$\mathrm{\frac{\pi}{6}}$

Option: 2
$\mathrm{\frac{\pi}{3}}$

Option: 3
$\mathrm{\frac{\pi}{2}}$

Option: 4
$\mathrm{\frac{\pi}{4}}$

Answers (1)

best_answer

The equation of the directrix is $\mathrm{x=\frac{a}{e} }$ and the focus is $\mathrm{S(a e, 0).}$

Let P be $\mathrm{(a \sec \theta, b \tan \theta)}$

Tangent at P is $\mathrm{\frac{x \sec \theta}{a}-\frac{y \tan \theta}{b}=1}$ $.......(1)$

At point of intersection $\mathrm{F, x=\frac{a}{e}}$ and $\mathrm{y=\frac{b(\sec \theta-e)}{e \tan \theta}}$

$\mathrm{ \begin{aligned} & \text { Then, slope of } F S \equiv m_1=\frac{\frac{b(\sec \theta-e)}{e \tan \theta}-0}{\frac{a}{e}-a e} \\ &=\frac{b(\sec \theta-e)}{-a \tan \theta\left(e^2-1\right)} \end{aligned} }$

And slope of $\mathrm{P S \equiv m_2}$

$\mathrm{ =\frac{b \tan \theta-0}{a \sec \theta-a e}=\frac{b \tan \theta}{a(\sec \theta-e)} }$

As $\mathrm{m_1 m_2=-1, angle \theta=\pi / 2}$

The answer is (c)

Posted by

himanshu.meshram

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