The slope of the tangent to y=y(x) at (x,y) is 2yx^2.

Question

The slope of the tangent to $y=y(x)$ at $(x,y)$ is $\dfrac{2y}{x^2}$. If the curve passes through the centre of $x^2+y^2-2x-2y=0$, its equation is:

Accepted Answer

$$\int\frac{dy}{y}=2\int\frac{dx}{x^2} \implies \ln|y|=-\frac2x+C.$$

Centre $(1,1)$: $0=-2+C \Rightarrow C=2$. So $\ln|y|=2-\dfrac2x \implies x\ln|y|=2(x-1)$.

Solutions · 1