Q96 / 160 · TS EAMCET 2019 (03 May Shift 1) · Mathematics
A tangent is drawn at $(\beta \sqrt{3} \cos \theta, \sin \theta)$ $\left(0 < \theta < \frac{\pi}{2}\right)$ 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
- Correct answerA$\frac{\pi}{6}$
- B$\frac{\pi}{3}$
- C$\frac{2 \pi}{3}$
- D$\frac{2 \pi}{4}$
Solution
Equation of tangent at $(3 \sqrt{3} \cos \theta, \sin \theta)$ on the ellipse $\frac{x^2}{27}+\frac{y^2}{1}=1$ is
$$
\begin{aligned}
& \frac{3 \sqrt{3} x \cos \theta}{27}+\frac{y \sin \theta}{1}=1 \\
& \frac{x}{3 \sqrt{3}} \cos \theta+\frac{y \sin \theta}{1}=1
\end{aligned}
$$
Sum of intercepts of tangent
i.e. $L=3 \sqrt{3} \sec \theta+\operatorname{cosec} \theta$
$\because \frac{d L}{d \theta}=3 \sqrt{3} \sec \theta \tan \theta-\operatorname{cosec} \theta \cot \theta$
For maxima or minima $\frac{d L}{d \theta}=0$
$3 \sqrt{3} \sec \theta \tan \theta-\operatorname{cosec} \theta \cot \theta=0$
$\tan ^3 \theta=\frac{1}{3 \sqrt{3}} \Rightarrow \tan \theta=1 \sqrt{3} \Rightarrow \theta=\frac{\pi}{6}$
Minimum at $\theta=\frac{\pi}{6}$
Hence, the correct answer is option: 'A'.