Question

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

The correct option is A $\frac{2}{\sqrt{3}}$

$x^2+3y^2=9$
equation of the normal at $(3\cos \theta ,\sqrt{3}\sin \theta )$
and $(-3\sin \theta ,\sqrt{3}\cos \theta )$ will be
$\frac{3x}{\cos \theta }-\frac{\sqrt{3}y}{\sin \theta }=6$
and $\frac{3x}{-\sin \theta }-\frac{\sqrt{3}y}{\cos \theta }=6$
Let $m_1,m_2$ be their slopes, then

$m_1=\sqrt{3}\tan \theta ,m_2=-\sqrt{3}\cot \theta$
So,
$\tan \beta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|\Rightarrow \tan \beta =\left|\frac{\sqrt{3}\tan \theta +\sqrt{3}\cot \theta }{1-\sqrt{3}\tan \theta \sqrt{3}\cot \theta }\right|\Rightarrow \tan \beta =\frac{\sqrt{3}}{2|\sin \theta \cos \theta |}\Rightarrow \tan \beta =\frac{\sqrt{3}}{\sin 2\theta }\Rightarrow \frac{1}{\cot \beta }=\frac{\sqrt{3}}{\sin 2\theta }\Rightarrow \frac{\cot \beta }{\sin 2\theta }=\frac{1}{\sqrt{3}}$
$\therefore \frac{2\cot \beta }{\sin 2\theta }=\frac{2}{\sqrt{3}}$