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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

 
Answers (1)
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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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