Question

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

• A. $(4,5)$
• B. $(5,4)$
• C. $(4,4)$
• D. $(5,5)$

Answer 1

The correct option is D $(5,5)$

$\frac{dy}{dx}=\frac{(x-2{)}^2+y+4}{(x-2)}=(x-2)+\frac{y+4}{(x-2)}$
Let $x-2=t\Rightarrow dx=dt$
and $y+4=u\Rightarrow dy=du\frac{dy}{dx}=\frac{du}{dt}\frac{du}{dt}=t+\frac{u}{t}\Rightarrow \frac{du}{dt}-\frac{u}{t}=tI.F.=e^{\int \frac{-1}{t}dt}=e^{-\ln t}=\frac{1}{t}u\times \frac{1}{t}=\int t\times \frac{1}{t}dt\Rightarrow \frac{u}{t}=t+c\frac{y+4}{x-2}=(x-2)+c$
Passing through $(0,0)$
$c=0$
$\Rightarrow (y+4)=(x-2{)}^2$
It passes though $(5,5)$