Let's consider the vertical parabola with the vertex at a generic point $$A(x_0,y_0)$$.
The focus is on $$F(x_0,y_0+\dfrac{p}{2})$$ and the generator line is given by the equation $$y=y_0-\dfrac{p}{2}$$.
The equation of the parabola is $$$(x-x_0)^2=2p(y-y_0)$$$
Given the parabola $$x^2-8y+16=0$$, find its focus, its vertex and the equation of its generator line.
First we should express the equation of the parabola in the form $$(x-x_0)^2=2p(y-y_0)$$.
To do this we can add $$8y-16$$ on both sides, and take $$8$$ as common factor: $$$x^2=8(y-2)$$$
Expressed as $$(x-0)^2=2\cdot4(y-2)$$ we obtain all the necessary information.
Then we can identify $$x_0=0, y_0=2, p=4$$.
The focus is on $$F(x_0,y_0+\dfrac{p}{2})$$, in this case $$F(0,4)$$.
The vertex is on $$A(x_0,y_0)$$ i.e. $$A(0,2)$$.
The equation of the generator line is $$y=y_0-\dfrac{p}{2}$$, in our case is $$y=0$$.