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

Answer 1

Answer: (B)

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

$x^2+3y^2=9$

$\Rightarrow 2x+6y\frac{dy}{dx}=0$ ....... Differentiating w.r.t $x$

$\Rightarrow \frac{dy}{dx}=\frac{-x}{3y}$

Equation of normal is

$-\frac{dx}{dy}=\frac{3y}{x}$

$\frac{dx}{dy}{|}_{(3\cos \theta ,\sqrt{3}\sin \theta )}=\frac{3\sqrt{3}\sin \theta }{-3\cos \theta }=\sqrt{3}\tan \theta =m_1$

$\frac{dx}{dy}{|}_{(-3\sin \theta ,\sqrt{3}\cos \theta )}=\frac{3\sqrt{3}\cos \theta }{-3\sin \theta }=-\sqrt{3}\cot \theta =m_2$

$\beta$ is the angle between the normals to the ellipse $\left(i\right)$, then

$\tan \beta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$=\left|\frac{\sqrt{3}\tan \theta +\sqrt{3}\cot \theta }{1-3\tan \theta \cot \theta }\right|$

$=\left|\frac{\sqrt{3}\tan \theta +\sqrt{3}\cot \theta }{1-3}\right|$

$\tan \beta =\frac{\sqrt{3}}{2}|\tan \theta +\cot \theta |$

$\frac{1}{\cot \beta }=\frac{\sqrt{3}}{2}|\tan \theta +\cot \theta |$

$\frac{1}{\cot \beta }=\frac{\sqrt{3}}{2}|\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }|$

$\frac{1}{\cot \beta }=\frac{\sqrt{3}}{2}\left|\frac{1}{\sin \theta \cos \theta }\right|$

$\frac{1}{\cot \beta }=\frac{\sqrt{3}}{\sin 2\theta }$

$\Rightarrow \frac{2\cot \beta }{\sin 2\theta }=\frac{2}{\sqrt{3}}$