# If β is one of the angles between the normals to the ellipse, $x^{2}+3y^{2}=9$ at the points     $\left ( -3\sin \theta ,\; \sqrt{3}\cos \theta \right );\; \; \theta \equiv \left ( 0,\frac{\pi }{2} \right )$   ; then $\frac{2\cot \beta }{\sin 2\theta }$ is equal to :  Option 1) $\frac{2}{\sqrt{3}}$       Option 2) $\frac{1}{\sqrt{3}}$ Option 3) ${\sqrt{2}}$ Option 4) $\frac{\sqrt{3}}{4}$

H Himanshu

As we learned

Angle between two lines (θ) -

$\tan \Theta = \frac{m_{2}-m_{1}}{1+m_{1}m_{2}}$

- wherein

Here $m_{1},m_{2}$ are the slope of two lines

$\tan \beta =\left | \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right |$

for tangent

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{-2x}{6y}=\frac{-x}{3y}$

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{-3\cos \theta }{\sqrt{3}\sin \theta \cdot 3}=\frac{-\cot \theta }{\sqrt{3}}$

Thus $m_{1}=\sqrt{3}\tan \theta$

Slope of normal  =  $\frac{3y}{x}$

= $-\sqrt{3}\cot \theta$

$\tan \beta =\left | \frac{\sqrt{3}\left ( \tan \theta +\cot \theta \right )}{1-3} \right |=\frac{\sqrt{3}}{\sin 2\theta }$

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

Option 1)

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

Option 2)

$\frac{1}{\sqrt{3}}$

Option 3)

${\sqrt{2}}$

Option 4)

$\frac{\sqrt{3}}{4}$

Exams
Articles
Questions