Question

If $\beta$ is one of the angles between the normals to the ellipse, $x^2 + 3y^2 = 9$ at the points $(3 \, \cos \theta , 3 \, \sin \theta )$ and $(-3 \, \sin \theta , 3 \, \cos \theta) ; \theta \epsilon \left( 0, \frac{\pi}{2} \right)$; then $\frac{2 \, \cot \, \beta}{\sin \, 2 \theta}$ is equal to

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\sqrt{2}$
• D. $\frac{\sqrt{3}}{4}$

Answer 1

Answer: (A)

$2 x d x+6 y d y=0$
$2 x d x=-c y d y$
$-\frac{ d x }{ dy }=\frac{3 y }{ x }$ =stepe of normal
$m _{1}=\frac{3 \sqrt{3} \sin \theta}{3 \cos \theta}=\sqrt{3} \tan \theta$
$m_{2}=\frac{3 \sqrt{3} \cos \theta}{-3 \sin \theta}-\sqrt{3} \cot \theta$
$\tan \beta=\left[\frac{\sqrt{3} \tan \theta+\sqrt{3} \cot \theta}{1+(\sqrt{3} \tan \theta)(-\sqrt{3} \cot \theta)}\right]$
$\tan \beta=\sqrt{3}\left(\mid \frac{\tan \theta+\cot \theta}{1-3}\right)$
$=\frac{\sqrt{3}}{2}(\tan \theta+\cot \theta)$
$\text { We have to find }$
$=\frac{2 \cot \beta}{\sin 2 \theta}$
$-\frac{22}{\sqrt{3}(\tan \theta+\cot \theta) 2 \sin \theta \cos \theta}$
$=\frac{2}{\sqrt{3}\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right) \sin \theta \cos \theta}$
$=\frac{2}{\sqrt{3}}$