Question

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

The correct option is 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 $\frac{xcos\theta }{3\sqrt{3}}+\frac{ysin\theta }{1}=1.$
Thus, sum of intercepts $=(\frac{3\sqrt{3}}{cos\theta }+\frac{1}{sin\theta })=f\left(\theta \right)$ [say]
$\Rightarrow f^{\prime }\left(\theta \right)=\frac{3\sqrt{3}si{n}^3\theta -co{s}^3\theta }{si{n}^2\theta co{s}^2\theta }$, put $f^{\prime }\left(\theta \right)=0$
$\Rightarrow si{n}^3\theta =\frac{1}{3^{3/2}}co{s}^3\theta$
$\Rightarrow tan\theta =\frac{1}{\sqrt{3}},i.e.\theta =\frac{\pi }{6}$ and at $\theta =\frac{\pi }{6},f^{\prime \prime }\left(0\right)>0$
Hence, tangent is minimum at $\theta =\frac{\pi }{6}$.