Downward parabola: value of p = 1/a?

[ducmod]

Homework Statement

Hello!
I am repeating conics, and have stumbled upon this paragraph on purplemath.com
"
Then a = –1/9. With a being the leading coefficient from the regular quadratic equation y = ax^2 + bx + c, I also know that the value of 1/a is the same as the value of 4p, so 1/(–1/9) = –9 = 4p, and thus p = –9/4. "

I am not aware of the rule that states that 4p equals 1/a for a parabola that has a vertex other than
(0, 0), which is the case here. The author refers to some general formula of parabola y = ax^2 + bx + c.
If vertex is at (0, 0 ), then the formula is x^2 = 4py, hence y = 1/4p * x^2, where a = 1/4p, hence
a = 1/4p. But in the given example, the vertex is not at (0, 0) - it's at (0, 25). Why a = 1/4p if the
vertex is not (0, 0)?

I would be grateful for explanation.

Thank you!

Homework Equations

The Attempt at a Solution

 

[pbuk]

Why a = 1/4p if the vertex is not (0, 0)?

This result can easily be shown for any parabola, not just one with vertex at the orgin e.g in this section of a Wikipedia page.