JEE PYQ: Application Of Derivatives Question 15
Question 15 - 2021 (24 Feb Shift 2)
For which of the following curves, the line $x + \sqrt{3}y = 2\sqrt{3}$ is the tangent at the point $\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$?
(1) $x^2 + 9y^2 = 9$
(2) $2x^2 - 18y^2 = 9$
(3) $y^2 = \frac{1}{6\sqrt{3}}x$
(4) $x^2 + y^2 = 7$
Show Answer
Answer: (1)
Solution
For $x^2 + 9y^2 = 9$: tangent at $\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$ is $\frac{3\sqrt{3}}{2} \cdot \frac{x}{9} + 9 \cdot \frac{1}{2} \cdot \frac{y}{9} = 1$, i.e., $\frac{\sqrt{3}x}{6} + \frac{y}{2} = 1 \Rightarrow x + \sqrt{3}y = 2\sqrt{3}$.
BETA
