Ellipse 2 Question 14
14. Tangent is drawn to ellipse $\frac{x^{2}}{27}+y^{2}=1$ at (3 $\sqrt{3} \cos \theta, \sin \theta)$ (where, $\theta \in(0, \pi / 2)$.
Then, the value of $\theta$ such that the sum of intercepts on axes made by this tangent is minimum, is
$(2003,1 M)$
(a) $\frac{\pi}{3}$
(b) $\frac{\pi}{6}$
(c) $\frac{\pi}{8}$
(d) $\frac{\pi}{4}$
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Answer:
Correct Answer: 14. (b)
Solution:
- 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 $\frac{x \cos \theta}{3 \sqrt{3}}+\frac{y \sin \theta}{1}=1$.
Thus, sum of intercepts $=\frac{3 \sqrt{3}}{\cos \theta}+\frac{1}{\sin \theta}=f(\theta) \quad$ [say]
$ \begin{aligned} & \Rightarrow \quad f^{\prime}(\theta)=\frac{3 \sqrt{3} \sin ^{3} \theta-\cos ^{3} \theta}{\sin ^{2} \theta \cos ^{2} \theta}, \text { put } f^{\prime}(\theta)=0 \\ & \Rightarrow \quad \sin ^{3} \theta=\frac{1}{3^{3 / 2}} \cos ^{3} \theta \\ & \Rightarrow \quad \tan \theta=\frac{1}{\sqrt{3}}, \text { i.e. } \theta=\frac{\pi}{6} \text { and at } \theta=\frac{\pi}{6}, f^{\prime \prime}(0)>0 \end{aligned} $
Hence, tangent is minimum at $\theta=\frac{\pi}{6}$.
