Given that the slope of the tangent to a curve $y=y(x)$ at any point $(x, y)$ is $\frac{2 y}{x^{2}}$. If the curve passes through the centre of the circle $x^{2}+y^{2}-2 x-2 y=0$, then its equation is:

(D) $x \log _{\mathrm{e}}|\mathrm{y}|=2(\mathrm{x}-1)$

(A) $x \log _{\mathrm{e}}|y|=x-1$

(C) $x^{2} \log _{e}|y|=-2(x-1)$

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Differential Equations

Given that the slope of the tangent to a curve

y = y (x)

at any point

(x, y)

\frac{2 y}{x ^{2}}

. If the curve passes through the centre of the circle

x^{2} + y^{2} - 2 x - 2 y = 0

, then its equation is:

Answer: (D)

The slope condition is a separable differential equation,

\frac{d y}{d x} = \frac{2 y}{x ^{2}}

, so

\frac{d y}{y} = \frac{2 d x}{x ^{2}}

. Integrating,

ln ∣ y ∣ = - \frac{2}{x} + C

. The curve passes through the centre of

x^{2} + y^{2} - 2 x - 2 y = 0

, which rewrites as

(x - 1)^{2} + (y - 1)^{2} = 2

with centre

(1, 1)

. Putting

(1, 1)

in,

ln 1 = - 2 + C

, so

C = 2

and

ln ∣ y ∣ = - \frac{2}{x} + 2 = \frac{2 ( x - 1 )}{x}

. Multiplying through by

x

, the curve is

x ln ∣ y ∣ = 2 (x - 1)

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option