Q

# Solve! - Differential equations - JEE Main-6

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

• Option 1)

$x\log_{e}\left | y \right |=2(x-1)$

• Option 2)

$x\log_{e}\left | y \right |=-2(x-1)$

• Option 3)

$x^{2}\log_{e}\left | y \right |=-2(x-1)$

• Option 4)

$x\log_{e}\left | y \right |=x-1$

Views

given slope of tangent $=\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2y}{x^{2}}$

$\frac{\mathrm{d} y}{\mathrm{y} }=\frac{\mathrm2{dx} }{\mathrm{x} ^{2}}$

Integrate but side.

$\ln \left ( y \right )=-\frac{2}{x}+C$

It passes through centre of circle $x^{2}+y^{2}-2x-2y=0$  $i.e\: \: \left ( 1,1 \right )$

$0=-2+C\Rightarrow C=2$

Eq. of curve is

$x\cdot \ln \left | y \right |=-2+2x$

$\Rightarrow x\cdot \ln \left | y \right |=2(x-1)$

Option 1)

$x\log_{e}\left | y \right |=2(x-1)$

Option 2)

$x\log_{e}\left | y \right |=-2(x-1)$

Option 3)

$x^{2}\log_{e}\left | y \right |=-2(x-1)$

Option 4)

$x\log_{e}\left | y \right |=x-1$

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