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. $L=3 \sqrt{3} \sec \,\theta+{\text {cosec}} \,\theta$
$\because \frac{d L}{d \theta}=3 \sqrt{3} \sec \,\theta \,\tan \,\theta-\text{cosec} \,\theta \cot \theta$
For maxima or minima $\frac{d L}{d \theta}=0$
$3 \sqrt{3} \sec \,\theta \,\tan \,\theta-\text{cosec}\, \theta \,\cot \,\theta=0$
$\tan ^{3} \theta=\frac{1}{3 \sqrt{3}} $
$\Rightarrow \tan \,\theta=1 \sqrt{3} $
$\Rightarrow \theta=\frac{\pi}{6}$
Minimum at $\theta=\frac{\pi}{6}$