Question

Tangent is drawn to ellipse $\frac{x^2}{27}+y^2=1$ at $(3\sqrt{3}\cos \theta ,\sin \theta )$

$($where $\theta \in (0,\pi /2$) $)$. Then the value of $\theta$ such that sum of intercepts on axes made by this tangent is minimum, is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{8}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (B)

$\frac{\pi }{6}$

Given, tangent is drawn at $(3\sqrt{3}\cos \theta ,\sin \theta )$ to $\frac{x^2}{27}+\frac{y^2}{1}=1$
Therefore equation of tangent is $\frac{x\cos \theta }{3\sqrt{3}}+\frac{y\sin \theta }{1}=1$
Thus, sum of intercept $=(\frac{3\sqrt{3}}{\cos \theta }+\frac{1}{\sin \theta })=f\left(\theta \right)$ (let)
$\Rightarrow f^{\prime }\left(\theta \right)=\frac{3\sqrt{3}{\sin }^3\theta -{\cos }^3\theta }{{\sin }^2\theta {\cos }^2\theta }$

Put $=f^{\prime }\left(\theta \right)=0$
$\Rightarrow {\sin }^3\theta =\frac{1}{3^{\frac{3}{2}}}{\cos }^3\theta \Rightarrow \tan \theta =\frac{1}{\sqrt{3}}\Rightarrow \theta =\frac{\pi }{6}$
and at $\theta =\frac{\pi }{6},f^{\prime \prime }\left(x\right)>0$
Therefore hence, tangent is minimum at $\theta =\frac{\pi }{6}$