JEE PYQ: Application Of Derivatives Question 58
Question 58 - 2019 (08 Apr Shift 2)
Given that the slope of the tangent to a curve $y = y(x)$ at any point $(x, y)$ is $\frac{2y}{x^2}$. If the curve passes through the centre of the circle $x^2 + y^2 - 2x - 2y = 0$, then its equation is:
(1) $x\log_e|y| = 2(x-1)$
(2) $x\log_e|y| = -2(x-1)$
(3) $x^2\log_e|y| = -2(x-1)$
(4) $x\log_e|y| = x - 1$
Show Answer
Answer: (1)
Solution
$\frac{dy}{dx} = \frac{2y}{x^2}$. Separating: $\frac{dy}{y} = \frac{2dx}{x^2}$. Integrating: $\ln|y| = -\frac{2}{x} + C$. Centre of circle $(1,1)$: $0 = -2 + C \Rightarrow C = 2$. So $x\ln|y| = 2(x-1)$.
BETA
