The slope of tangent at any point x ,   y on a curve y = y x is x 2 + y 2 2 x y ,   x > 0 . If…

Given:

$\frac{dy}{dx}=\frac{{\displaystyle x^2+y^2}}{{\displaystyle 2xy}}$

Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$v+x\frac{dv}{dx}=\frac{{\displaystyle 1+v^2}}{{\displaystyle 2v}}$

$\Rightarrow x\frac{dv}{dx}=\frac{{\displaystyle 1-v^2}}{{\displaystyle 2v}}$

$\Rightarrow \int \left(\frac{2v}{v^2-1}\right)dv=-\int \frac{dx}{x}$

$\Rightarrow {\log }_e\left|v^2-1\right|={\log }_e\left(\frac{C}{x}\right)$

$\Rightarrow \frac{y^2-x^2}{x^2}=\frac{C}{x}$

$\Rightarrow y^2-x^2=Cx$

Put $x=2$ and $y=0$ we get,

$0-2^2=2C\Rightarrow C=-2$

$\Rightarrow y^2=x^2-2x$

$\Rightarrow y\left(8\right)=\sqrt{8^2-16}$

$\Rightarrow y\left(8\right)=\sqrt{48}=4\sqrt{3}$