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

Let the slope of tangent at any point <br/><br/>$(x, y)$ on a curve $y=y(x)$ is $\frac{d y}{d x}$ <br/><br/>According to the question, $\frac{d y}{d x}=\frac{x^2+y^2}{2 x y}(x>0)$ [Given] <br/><br/>$$ \begin{aligned} &\text { Let } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\\\ &\Rightarrow v+x \frac{d v}{d x}=\frac{x^2+v^2 x^2}{2 v x^2}=\frac{1+v^2}{2 v} \\\\ &\Rightarrow x \frac{d v}{d x}=\frac{1+v^2}{2 v}-v=\frac{1-v^2}{2 v} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow \int \frac{2 v d v}{1-v^2}=\int \frac{d x}{x} \text { [Taking integration on both sides] } \\\\ & \Rightarrow-\ln \left|1-v^2\right|=\ln x-\ln c \\\\ & \Rightarrow \ln \left(1-v^2\right) x=\ln c \quad(c=\text { constant }) \\\\ & \Rightarrow \left(1-y^2 / x^2\right) \cdot x=c \\\\ & \Rightarrow \left(x^2-y^2\right)=c x \end{aligned} $$ <br/><br/>Put $y(2)=0 \Rightarrow c=2$ <br/><br/>$$ \begin{array}{rlrl} & \therefore x^2-y^2 =2 x \\\\ & \therefore y^2 =x^2-2 x \\\\ & \Rightarrow y(x) = \pm \sqrt{x^2-2 x} \end{array} $$ <br/><br/>Put $x=8$, we get <br/><br/>$\Rightarrow y(8)= \pm \sqrt{64-16}= \pm \sqrt{48}= \pm 4 \sqrt{3}$