Let y=y(x) with y>0 be the solution of (x^2+y^2 dx=2xy dy for x>0 passing through the point (2,√2).

The equation is homogeneous; put $y=vx$. Then $(1+v^2)=2v\left(v+x\frac{dv}{dx}\right)$ gives $2vx\,dv=(1-v^2)\,dx$, hence $\frac{2v\,dv}{1-v^2}=\frac{dx}{x}$ and $-\ln|1-v^2|=\ln x+C$, i.e. $x^2-y^2=cx$. The point $(2,\sqrt2)$ gives $c=1$, so $y^2=x^2-x$. At $x=4$, $y=\sqrt{12}=2\sqrt3$: (A) true. Writing it as $\left(x-\frac12\right)^2-y^2=\frac14$ shows a rectangular hyperbola, eccentricity $\sqrt2$, not $2$: (B) false. Differentiating, $2yy'=2x-1$, so at $(2,\sqrt2)$ the slope is $\frac{3}{2\sqrt2}=\frac{3\sqrt2}{4}\ne\frac34$: (C) false. The tangent meets the $x$-axis where $x=2-\sqrt2\cdot\frac{2\sqrt2}{3}=\frac23$, not $\frac43$: (D) false. Hence only option A is correct. The homogeneous substitution $y=vx$ and the observation that the family $x^2-y^2=cx$ is a set of rectangular hyperbolas through the origin make each remaining option a quick check.