Pick a point on the plane. Find the equation of the vertical parabola whose focus is the chosen point and has $$p=14$$.
See development and solution
Development:
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 $$$(x-5)^2=28(y+2)$$$
Solution:
Taking the point $$F(5,5)$$ we obtain the parabola $$(x-5)^2=28(y+2)$$.