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If the line $y\, = \,mx\, + \,7\sqrt 3 $ is normal to the hyperbola $\frac{{{x^2}}}{{24}} - \frac{{{y^2}}}{{18}} = 1,$ then a value of $m$ is
$\frac{2}{{\sqrt 5 }}$
$\frac{{\sqrt 5 }}{2}$
$\frac{{\sqrt {15} }}{2}$
$\frac{3}{{\sqrt 5 }}$
Solution
$\frac{{{x^2}}}{{24}} – \frac{{{y^2}}}{{18}} = 1\,\,\,\,\, \Rightarrow a\, = \sqrt {24} :b\, = \sqrt {18} $
Paramentric normal:
$\sqrt {24} \cos \theta .x + \sqrt {18} .y\cot \theta = 42$
At $x = 0;y = \frac{{42}}{{\sqrt {18} }}\tan \theta = 7\sqrt 3 $ (from given equation)
$ \Rightarrow \tan \theta = \sqrt {\frac{3}{2}} \Rightarrow \sin \theta = \pm \sqrt {\frac{3}{5}} $
slope of parametric normal $ = \frac{{ – \sqrt {24} \cos \theta }}{{\sqrt {18} \cot \theta }} = m$
$ \Rightarrow m = – \sqrt {\frac{4}{3}} \sin \theta = – \frac{2}{{\sqrt 5 }}$
