#### If $\beta$  is one of the angles between the normals to the ellipse,  $\mathrm {x^2+3 y^2=9}$  at the points  $\mathrm {(3 \cos \theta, \sqrt{3} \sin \theta)}$  and $\mathrm {(-3 \sin \theta, \sqrt{3} \cos \theta); \theta \in\left(0, \frac{\pi}{2}\right), if\: \: \frac{2 \cot \beta}{\sin 2 \theta}}$   is equal to   $\mathrm {\frac{m}{\sqrt{n}}, then\: \: m+n}$  isOption: 1 1Option: 2 3Option: 3 5Option: 4 7

Given, $\mathrm {\frac{x^2}{9}+\frac{y^2}{3}=1}$
Equation of normal at  $\mathrm {P(3 \cos \theta, \sqrt{3} \sin \theta) }$  is

$\mathrm { 3 \sec \theta x-\sqrt{3} \operatorname{cosec} \theta y=6 }......(\mathrm i)$
Slope of (i) is given by $\mathrm { \frac{-3 \sec \theta}{-\sqrt{3} \operatorname{cosec} \theta}=\sqrt{3} \tan \theta }$

Equation of normal at  $\mathrm { (-3 \sin \theta, \sqrt{3} \cos \theta) }$   is,

$\mathrm { -3 \operatorname{cosec} \theta x-\sqrt{3} \sec \theta y=6 }......(\mathrm {ii})$

Slope of (ii) is given by  $\frac{-(-3 \operatorname{cosec} \theta)}{(-\sqrt{3} \sec \theta)}=-\sqrt{3} \cot \theta$

Angle between normals is  $\beta$
\begin{aligned} \therefore \quad & \tan \beta=\left|\frac{\sqrt{3} \tan \theta+\sqrt{3} \cot \theta}{1-3}\right|=\left|\frac{-\sqrt{3}}{2}\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)\right| \\ & =\left|\frac{-\sqrt{3}\left(\sin ^2 \theta+\cos ^2 \theta\right)}{2 \sin \theta \cos \theta}\right|=\frac{\sqrt{3}}{\sin 2 \theta} \end{aligned}

\begin{aligned} & \Rightarrow \tan \beta=\frac{\sqrt{3}}{\sin 2 \theta} \Rightarrow \cot \beta=\frac{\sin 2 \theta}{\sqrt{3}} \\ & \therefore\mathrm { \quad \frac{2 \cot \beta}{\sin 2 \theta}=\frac{2}{\sqrt{3}}=\frac{m}{\sqrt{n}}, \Rightarrow m+n=5} \end{aligned}