Given that the slope of the tangent to a curve y = y(x) at any point (x,y) is (2y/x2). If the curve passes through the centre of the circle x2 + y2 - 2x - 2y = 0, then its equation is :

Question

Given that the slope of the tangent to a curve y = y(x) at any point (x,y) is (2y/x2). If the curve passes through the centre of the circle x2 + y2 - 2x - 2y = 0, then its equation is : (A) x loge |y| = 2 (x - 1) (B)  x loge |y| = x - 1 (C) x2 loge |y| = - (x - 1) (D) x loge |y| = - 2(x - 1)

Tardigrade Admin · Accepted Answer

given (dy/dx) = (2y/x2) ⇒ ∫ (dy/2y) = ∫ (dx/x2) ⇒ (1/2) ℓ ny =- (1/x) + c passes through centre (1,1) ⇒ c = 1 ⇒ x ℓ ny = 2(x-1)

Q. 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 :

$x lo g_{e} ∣ y ∣ = 2 (x - 1)$

$x lo g_{e} ∣ y ∣ = x - 1$

$x^{2} lo g_{e} ∣ y ∣ = - (x - 1)$

$x lo g_{e} ∣ y ∣ = - 2 (x - 1)$

given $\frac{d y}{d x} = \frac{2 y}{x ^{2}}$
$\Rightarrow \int \frac{d y}{2 y} = \int \frac{d x}{x ^{2}}$
$\Rightarrow \frac{1}{2} ℓ n y = - \frac{1}{x} + c$
passes through centre (1,1)
$\Rightarrow c = 1$
$\Rightarrow x ℓ n y = 2 (x - 1)$

Q. Given that the slope of the tangent to a curve y=y(x) at any point (x,y) is x22y​. If the curve passes through the centre of the circle x2+y2−2x−2y=0, then its equation is :

xloge​∣y∣=2(x−1)

xloge​∣y∣=x−1

x2loge​∣y∣=−(x−1)

xloge​∣y∣=−2(x−1)