A geometry problem by vishnu c

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

$put~y=vx$

and solving

$ln|1-v^{2}|=ln(\frac{c}{x})$ , where c is constant of integration

so the curve is $x^{2}-y^{2}=cx$

No need of finding $c$ as it is a rectangular hyperbola as clear from the equation

$\huge{\frac{(x-\frac{c}{2})^{2}}{(\frac{c}{2})^{2}}-\frac{y^{2}}{(\frac{c}{2})^{2}}=1}$

and eccentricity of rectangular hyperbola is always $\large{\boxed{\sqrt{2}}}$