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If the tangent on the point $(2\sec \phi ,\;3\tan \phi )$ of the hyperbola $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{9} = 1$ is parallel to $3x - y + 4 = 0$, then the value of $\phi$ is ............ $^o$
$45$
$60$
$30$
$75$
Solution
(c) Differentiation of $x = 2\sec \phi $
==> $\frac{{dx}}{{d\phi }} = 2\sec \phi \tan \phi $
Differentiate, $y = 3\tan \phi $ w.r.t. $\theta$,
we get $\frac{{dy}}{{d\phi }} = 3{\sec ^2}\phi $
$\therefore $Gradient of tangent $\frac{{dy}}{{dx}} = \frac{{dy/d\phi }}{{dx/d\phi }} = \frac{{3{{\sec }^2}\phi }}{{2\sec \phi \tan \phi }}$
$\frac{{dy}}{{dx}} = \frac{3}{2}\,{\rm{cosec}}\phi $ …..$(i)$
But, tangent is parallel to $3x – y + 4 = 0$
$\therefore $Gradient $ m = 3$ …..$(ii)$
By $(i)$ and $(ii),$ $\frac{3}{2}{\rm{cosec}}\phi = 3$
==> ${\rm{cosec}}\phi = 2$,
$\therefore \phi = 30^\circ $.
