The slope of tangent at any point (x,y) on a curve y=y(x) is x^2+y^2 2xy,x>0. If y(2)=0, then a value of y(8) is

Given:

$$\frac{dy}{dx}=\frac{{x}^{2}+{y}^{2}}{2xy}$$dxdy​=2xyx2+y2​

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

$$v+x\frac{dv}{dx}=\frac{1+{v}^{2}}{2v}$$v+xdxdv​=2v1+v2​

$$\Rightarrow x\frac{dv}{dx}=\frac{1-{v}^{2}}{2v}$$⇒xdxdv​=2v1−v2​

$$\Rightarrow \int (\frac{2v}{{v}^{2}-1})dv=-\int \frac{dx}{x}$$⇒∫(v2−12v​)dv=−∫xdx​

$$\Rightarrow {\mathrm{log}}_{e}|{v}^{2}-1|={\mathrm{log}}_{e}(\frac{C}{x})$$⇒loge​∣v2−1∣=loge​(xC​)

$$\Rightarrow \frac{{y}^{2}-{x}^{2}}{{x}^{2}}=\frac{C}{x}$$⇒x2y2−x2​=xC​

$$\Rightarrow {y}^{2}-{x}^{2}=Cx$$⇒y2−x2=Cx

Put $x=2$x=2 and $y=0$y=0 we get,

$$0-{2}^{2}=2C\Rightarrow C=-2$$0−22=2C⇒C=−2

$$\Rightarrow {y}^{2}={x}^{2}-2x$$⇒y2=x2−2x

$$\Rightarrow y(8)=\sqrt{{8}^{2}-16}$$⇒y(8)=82−16​

$$\Rightarrow y(8)=\sqrt{48}=4\sqrt{3}$$⇒y(8)=48​=43​