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The value of $m$, for which the line $y = mx + \frac{{25\sqrt 3 }}{3}$, is a normal to the conic $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$, is
$\sqrt 3 $
$ - \frac{2}{{\sqrt 3 }}$
$ - \frac{{\sqrt 3 }}{2}$
$1$
Solution
(b) We know that the equation of the normal of the conic ,
$\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ at point $(a\sec \theta ,\,b\tan \theta )$
is $ax\sec \theta + by\cot \theta = {a^2} + {b^2}$
or $y = \frac{{ – a}}{b}\sin \theta \,x + \frac{{{a^2} + {b^2}}}{{b\cot \theta }}$
Comparing above equation with equation $y = mx + \frac{{25\sqrt 3 }}{3}$ and taking $a = 4,\,b = 3$
we get, $\frac{{{a^2} + {b^2}}}{{b\cot \theta }} = \frac{{25\sqrt 3 }}{3}$
$ \Rightarrow $ $\tan \theta = \sqrt 3$
$\Rightarrow \theta = {60^o}$
and $m = – \frac{a}{b}\sin \theta = \frac{{ – 4}}{3}\sin {60^o}$
= $\frac{{ – 4}}{3} \times \frac{{\sqrt 3 }}{2} = \frac{{ – 2}}{{\sqrt 3 }}$.
