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

The slope of the tangent $$\frac{dy}{dx}$$ is given as:

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

This is a homogeneous differential equation. We use the substitution $$y = vx$$, which implies $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$.

$$v + x\frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{1 + v^2}{2v}$$

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

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

Using $$u = 1 - v^2$$, $$du = -2v \, dv$$:

$$-\ln|1 - v^2| = \ln|x| + \ln|C|$$

$$\ln\left|\frac{1}{1 - v^2}\right| = \ln|Cx| \implies \frac{1}{1 - v^2} = Cx$$

Substitute $$v = \frac{y}{x}$$:

$$\frac{1}{1 - \frac{y^2}{x^2}} = Cx \implies \frac{x^2}{x^2 - y^2} = Cx \implies \mathbf{x^2 - y^2 = \frac{x}{C}}$$

Let $$K = \frac{1}{C}$$, then $$x^2 - y^2 = Kx$$.

$$2^2 - 0^2 = K(2) \implies 4 = 2K \implies K = 2$$

The equation of the curve is $$x^2 - y^2 = 2x$$.

Substitute $$x = 8$$:

$$8^2 - y^2 = 2(8)$$

$$64 - y^2 = 16$$

$$y^2 = 48 \implies y = \pm \sqrt{48} = \pm 4\sqrt{3}$$

Since the options provide the positive value:

$$y(8) = 4\sqrt{3}$$

Correct Option: (D)