Equation of the vertical parabola with generic vertex
Pick a point on the plane. Find the equation of the vertical parabola whose focus is the chosen point and has $p=14$.
Choosing $F(5,5)$ and identifying with $F(x_0,y_0+\dfrac{p}{2})$ we have $x_0=5$ and $y_0+\dfrac{p}{2}=5$. We can also compute $y_0=5-\dfrac{14}{2}=5-7=-2$.
We have the necessary elements of the equation. Substituting in $(x-x_0)^2=2p(y-y_0)$ we obtain $$(x-5)^2=28(y+2)$$
Taking the point $F(5,5)$ we obtain the parabola $(x-5)^2=28(y+2)$.