Question

Given that the slope of the tangent to a curve $y=y\left(x\right)$ 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 :

• A. $x{\log }_e|y|=2(x-1)$
• B. $x{\log }_e|y|=(x-1)$
• C. $x{\log }_e|y|=-2(x-1)$
• D. $x^2{\log }_e|y|=-2(x-1)$

Answer 1

The correct option is A $x{\log }_e|y|=2(x-1)$

$\frac{dy}{dx}=\frac{2y}{x^2}$
$\Rightarrow \frac{dy}{y}=\frac{2dx}{x^2}$
Integrating both sides
$\int \frac{dy}{y}=\int \frac{2dx}{x^2}$
$\Rightarrow {\log }_e|y|=\frac{-2}{x}+c$

The curve passes through the centre of the circle $x^2+y^2-2x-2y=0$, i.e., $(1,1)$
$\Rightarrow 0=-2+c$
$\Rightarrow c=2$
$\therefore$ Equation of the curve is
${\log }_e|y|=\frac{-2}{x}+2$
$\Rightarrow x{\log }_e|y|=2(x-1)$