Given that the slope of the tangent to a curve

Question:

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 :

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

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

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

  4. (4) $x \log _{e}|y|=x-1$


Correct Option: 1,

Solution:

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

Integrating both sides, $\int \frac{d y}{y}=2 \int \frac{d x}{x^{2}}$

$\Rightarrow \ln |y|=-\frac{2}{x}+C$ .........(1)

Equation (i) passes through the centre of the circle

$x^{2}+y^{2}-2 x-2 y=0$, i.e., $(1,1)$

$\therefore C=2$

Now, $\ln |y|=-\frac{2}{x}+2$

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

 

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