• question_answer 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 [JEE Main 9-4-2019 Morning] A) $\frac{\sqrt{5}}{2}$                         B) $\frac{3}{\sqrt{5}}$C) $\frac{2}{\sqrt{5}}$             D) $\frac{\sqrt{15}}{2}$

$\frac{{{x}^{2}}}{24}-\frac{{{y}^{2}}}{18}=1\Rightarrow a=\sqrt{24};b=\sqrt{18}$ Parametric 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$=\frac{-\sqrt{24}\cos \theta }{\sqrt{18}\cot \theta }=m$ $\Rightarrow m=-\sqrt{\frac{4}{3}}\sin \theta =-\frac{2}{\sqrt{5}}or\frac{2}{\sqrt{5}}$