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Q.
If a curve passes through the origin and the slope of the tangent to it at any point $( x , y )$ is $\frac{x^{2}-4 x+y+8}{x-2}$, then this curve also passes through the point:
Given
$y (0)=0$
$\& \frac{d y}{d x}=\frac{(x-2)^{2}+y+4}{x-2}$
$\Rightarrow \frac{ dy }{ dx }-\frac{ y }{ x -2}=( x -2)+\frac{4}{ x -2}$
$\Rightarrow I.F. = e ^{-\int \frac{1}{ x -2} dx }=\frac{1}{ x -2}$
Solution of $L.D.E$
$\Rightarrow y \frac{1}{x-2}=\int \frac{1}{x-2}\left((x-2)+\frac{4}{x-2}\right) \cdot d x$
$\Rightarrow \frac{y}{x-2}=x-\frac{4}{x-2}+C$
Now, at $x=0, y=0 \Rightarrow C=-2$
$y=x(x-2)-4-2(x-2)$
$\Rightarrow y=x^{2}-4 x$
This curve passes through (5,5)