Question

Tangent is drawn to ellipse $\frac{x^2}{27}+\frac{y^2}{1}=1,(3\sqrt{3}\cos \theta ,\sin \theta )$, where $\left(\theta ϵ(0,\frac{\pi }{2})\right)$. Then the value of $\theta$ such that the sum of intercept 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

The correct option is B $\frac{\pi }{6}$

Equation of tangent is drawn at a $(3\sqrt{3}\cos \theta ,\sin \theta )$ to the curve
$\frac{x^2}{27}+\frac{y^2}{1}=1$ is $\frac{x\cos \theta }{3\sqrt{3}}+\frac{y\sin \theta }{1}=1$
Thus, sum of intercepts
$=(3\sqrt{3}\sec \theta +cosec\theta )=f\left(\theta \right)\left(say\right)$
$\Rightarrow f^{\prime }\left(\theta \right)=3\sqrt{3}\sin \theta \tan \theta -cosec\theta \cot \theta$
$=\frac{3\sqrt{3}{\sin }^3\theta -{\cos }^3\theta }{{\sin }^2\theta {\cos }^2\theta }$
Put $f^{\prime }\left(\theta \right)=0$
$\Rightarrow {\sin }^3\theta =\frac{1}{3^{3/2}}{\cos }^3\theta$
$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}$
$\Rightarrow \theta =\frac{\pi }{6}$.