If a curve passes through the origin and the slope of the tangent to it at any point $\left(x,y\right)$ is  $\frac{x^2-4x+y+8}{x-2}$ then this curve also passes through the point

$\begin{array}{l}\frac{dy}{dx}=\frac{x^2-4x+y+8}{x-2}=\frac{(x-2{)}^2+(y+4)}{(x-2)}\\ \frac{dy}{dx}=(x-2)+\frac{(y+4)}{(x-2)}\\ \mid \text{ plug in }x-2=t\Rightarrow dx=dt\\ y+4=u\Rightarrow dy=du,\frac{dy}{dx}=\frac{du}{dt}\\ \Rightarrow \frac{du}{dt}=t+\frac{u}{t}\Rightarrow \frac{du}{dt}-\frac{u}{t}=t\\ \text{ I.F }=e^{\int -\frac{1}{t}dt}=e^{-\ln \left(t\right)}=\frac{1}{t}\end{array}$

$\begin{array}{l}\text{ Solution is, }u·\frac{1}{t}=\int t·\frac{1}{t}dt\Rightarrow \frac{u}{t}=t+c\\ \frac{y+4}{x-2}=x-2+c\\ \text{ Passing through }(0,0)\Rightarrow c=0\\ \Rightarrow y+4=(x-2{)}^2\end{array}$