If the tangent on the point (2sec phi ,;3tan | vedclass

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Question

(c) Differentiation of $x = 2\sec \phi $

==> $\frac{{dx}}{{d\phi }} = 2\sec \phi 	an \phi $

Differentiate, $y = 3	an \phi $  w.r.t. $\

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(c) Differentiation of $x = 2\sec \phi $

==> $\frac{{dx}}{{d\phi }} = 2\sec \phi 	an \phi $

Differentiate, $y = 3	an \phi $  w.r.t. $	heta$,

we get $\frac{{dy}}{{d\phi }} = 3{\sec ^2}\phi $

$	herefore $Gradient of tangent $\frac{{dy}}{{dx}} = \frac{{dy/d\phi }}{{dx/d\phi }} = \frac{{3{{\sec }^2}\phi }}{{2\sec \phi 	an \phi }}$

$\frac{{dy}}{{dx}} = \frac{3}{2}\,{m{cosec}}\phi $ .....$(i)$

But, tangent is parallel to $3x - y + 4 = 0$

$	herefore $Gradient $ m = 3$ .....$(ii)$

By $(i)$ and $(ii),$ $\frac{3}{2}{m{cosec}}\phi = 3$

==> ${m{cosec}}\phi = 2$,

$	herefore \phi = 30^\circ $.

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 ............ $^o$

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