$y^2=16x$ describes a parabola opening to the right, not a hyperbola. The vertex is at the origin. So the tangents with positive slope will be the ones that lie above the $x$-axis (those corresponding to positive values of $y$), and the tangents with negative slope will be those that lie below the $x$-axis (negative values of $y$).
I note that you can do this without having to figure all of that out, and without having to "solve for $y$", by using implicit differentiation. From $y^2 = 16x$ we can take derivatives on both sides, using the Chain Rule for $y$ to get: $\begin{align*} y^2 &= 16x\\ \frac{d}{dx} y^2 = \frac{d}{dx} 16x\\ 2yy' &= 16\\ y' &= \frac{16}{2y}\\ y' &= \frac{8}{y}. \end{align*}$ So, provided we are not at the point $(0,0)$ (where the tangent is vertical), the slope of the tangent will be $\frac{8}{y}$ at the point $(x,y)$ (where $y^2 = 16x$). This will automatically take care of the "sign".