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 $\frac{2}{\sqrt{3}}$

$x^2+3y^2=9$
Differentiating w.r.t. $x$,
$2x+6y{y}^{\prime }=0\Rightarrow y^{\prime }=-\frac{x}{3y}$
Slope of the normal,
$m=\frac{3y}{x}$
Slope of the normal at $(3\cos \theta ,\sqrt{3}\sin \theta )$
$m_1=3\frac{\sqrt{3}\sin \theta }{3\cos \theta }=\sqrt{3}\tan \theta$
Slope of the normal at $(-3\sin \theta ,\sqrt{3}\cos \theta )$
$m_2=3\frac{\sqrt{3}\cos \theta }{-3\sin \theta }=-\sqrt{3}\cot \theta$
So,
$\tan \beta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|\Rightarrow \tan \beta =\left|\frac{\sqrt{3}\tan \theta +\sqrt{3}\cot \theta }{1-\sqrt{3}\tan \theta \sqrt{3}\cot \theta }\right|\Rightarrow \tan \beta =\frac{\sqrt{3}}{2|\sin \theta \cos \theta |}\Rightarrow \tan \beta =\frac{\sqrt{3}}{\sin 2\theta }\Rightarrow \frac{1}{\cot \beta }=\frac{\sqrt{3}}{\sin 2\theta }\Rightarrow \frac{\cot \beta }{\sin 2\theta }=\frac{1}{\sqrt{3}}$
$\therefore \frac{2\cot \beta }{\sin 2\theta }=\frac{2}{\sqrt{3}}$