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  • If   is one of the angles between the normals to the ellipse,  <img alt="\mathrm {x^2+3 y^2=9}" src="https

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}  is

Option: 1

1


Option: 2

3


Option: 3

5


Option: 4

7


Answers (1)

best_answer

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}

 

Posted by

Gautam harsolia

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