Mathematics - Ellipse Question with Solution | TestHub

Given equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{3}=1$

Normal at $\left(3\cos \theta ,\sqrt{3}\sin \theta \right)$ is

$3\sec \theta .x-\sqrt{3}\mathrm{cosec}\theta y=6$

$\Rightarrow$Slope $m_1=\frac{3\sec \theta }{\sqrt{3}\mathrm{cosec}\theta }=\sqrt{3}\tan \theta$

Normal at $\left(-3\mathrm{sin\theta },\sqrt{3}\cos \theta \right)$

$-3\mathrm{cosec}\theta .x-\sqrt{3}\sec \theta .y=6$

$\Rightarrow$Slope $m_2=\frac{-3\mathrm{cosec}\theta }{\sqrt{3}\sec \theta }=-\sqrt{3}\cot \theta$

Angle between normal is $\beta$.

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

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