Equation of the vertical parabola with generic vertex

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 $y=0$.