Question

A tangent is drawn at the point $(3\sqrt{3}\cos \theta ,\sin \theta )$ for 0< \theta < \pi/2 of an ellipse $\frac{x^2}{27}+\frac{y^2}{1}=1$ the least value of the sum of the intercepts on the coordinates axis by tangent is attained at $\theta$ equal to ?

Answer 1

Equation of the tangent at $(3\sqrt{3}\cos \theta ,\sin \theta )$ is

$\frac{x\left(3\sqrt{3}\cos \theta \right)}{27}+\frac{y\left(\sin \theta \right)}{1}=1$

$\Rightarrow \frac{x\left(3\sqrt{3}\cos \theta \right)}{27}+y\sin \theta =1$

$\therefore$Sum of intercepts on axes$=3\sqrt{3}\sec \theta +\csc \theta =f\left(\theta \right)$

$\therefore f^{\prime }\left(\theta \right)=3\sqrt{3}\sec \theta \tan \theta -\csc \theta \cot \theta =0$

$\Rightarrow \frac{3\sqrt{3}}{\cos \theta }\frac{\sin \theta }{\cos \theta }-\frac{1}{\sin \theta }\frac{\cos \theta }{\sin \theta }=0$

$\Rightarrow \frac{3\sqrt{3}\sin \theta }{{\cos }^2\theta }-\frac{\cos \theta }{{\sin }^2\theta }=0$

$\Rightarrow \frac{3\sqrt{3}\sin \theta }{{\cos }^2\theta }=\frac{\cos \theta }{{\sin }^2\theta }$

$\Rightarrow 3\sqrt{3}{\sin }^3\theta ={\cos }^3\theta$

$\Rightarrow {\tan }^3\theta =\frac{1}{3\sqrt{3}}$

$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}$

$\therefore \theta =\frac{\pi }{6}$