JEE PYQ: Application Of Derivatives Question 8
Question 8 - 2021 (18 Mar Shift 2)
Let a tangent be drawn to the ellipse $\frac{x^2}{27} + y^2 = 1$ at $(3\sqrt{3}\cos\theta, \sin\theta)$ where $\theta \in \left(0, \frac{\pi}{2}\right)$. Then the value of $\theta$ such that the sum of intercepts on axes made by this tangent is minimum is equal to:
(1) $\frac{\pi}{8}$
(2) $\frac{\pi}{4}$
(3) $\frac{\pi}{6}$
(4) $\frac{\pi}{3}$
Show Answer
Answer: (3)
Solution
Tangent: $\frac{x\cos\theta}{3\sqrt{3}} + y\sin\theta = 1$. Sum of intercepts $= 3\sqrt{3}\sec\theta + \csc\theta$. Setting derivative to zero: $3\sqrt{3}\sec\theta\tan\theta - \csc\theta\cot\theta = 0 \Rightarrow \tan^2\theta = \frac{1}{3\sqrt{3}} \cdot \frac{1}{\sqrt[3]{1}}$. This gives $\theta = \frac{\pi}{6}$.
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