The slope of tangent at any point (x, y) on a curve y = y(x) is x ^...

Given:

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

Put $y = vx ⇒ \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$

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

$⇒ x\dfrac{dv}{dx} = \dfrac{1 - {v}^{2}}{2v}$

$⇒ ∫ \left(\dfrac{2v}{{v}^{2} - 1}\right)dv = - ∫ \dfrac{dx}{x}$

$⇒ \log_{e}\left|{v}^{2} - 1\right| = \log_{e}\left(\dfrac{C}{x}\right)$

$⇒ \dfrac{{y}^{2} - {x}^{2}}{{x}^{2}} = \dfrac{C}{x}$

$⇒ {y}^{2} - {x}^{2} = Cx$

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

$0 - {2}^{2} = 2C ⇒ C = - 2$

$⇒ {y}^{2} = {x}^{2} - 2x$

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

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