Question

A tangent is drawn at $(3\sqrt{3}\cos \theta ,\sin \theta )$ $\left(0<\theta <\frac{\pi }{2}\right)$ to the ellipse $\frac{x^2}{27}+\frac{y^2}{1}=1$. The value of $\theta$ for which the sum of the intercepts on the coordinate axes made by this tangent attains the minimum, is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{2\pi }{3}$
• D. $\frac{2\pi }{4}$

Answer 1

Answer: (A)

Equation of tangent at $(3\sqrt{3}\cos \theta ,\sin \theta )$ on the
ellipse $\frac{x^2}{27}+\frac{y^2}{1}=1$ is
$\frac{3\sqrt{3}x\cos \theta }{27}+\frac{y\sin \theta }{1}=1$
$\frac{x}{3\sqrt{3}}\cos \theta +\frac{y\sin \theta }{1}=1$
Sum of intercepts of tangent
i.e. $L=3\sqrt{3}\sec \theta +\text{cosec}\theta$
$\because \frac{dL}{d\theta }=3\sqrt{3}\sec \theta \tan \theta -\text{cosec}\theta \cot \theta$
For maxima or minima $\frac{dL}{d\theta }=0$
$3\sqrt{3}\sec \theta \tan \theta -\text{cosec}\theta \cot \theta =0$
${\tan }^3\theta =\frac{1}{3\sqrt{3}}$
$\Rightarrow \tan \theta =1\sqrt{3}$
$\Rightarrow \theta =\frac{\pi }{6}$
Minimum at $\theta =\frac{\pi }{6}$