check the picture below.
since the focus point is above the directrix, that simply means the parabola is a vertical one, and therefore the square variable is the "x".
keeping in mind that, there's a distance "p" from the vertex to either the focus point or the directrix, that puts the vertex half-way between those fellows, in this case at 0, right between 1 and -1, as you see in the picture, and the parabola looks like so.
since the parabola is opening upwards, the value for "p" is positive, thus
[tex]\bf \textit{parabola vertex form with focus point distance}
\\\\
\begin{array}{llll}
4p(x- h)=(y- k)^2
\\\\
\boxed{4p(y- k)=(x- h)^2}
\end{array}
\qquad
\begin{array}{llll}
vertex\ ( h, k)\\\\
p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
h=5\\
k=0\\
p=1
\end{cases}\implies 4(1)(y-0)=(x-5)^2
\\\\\\
4y=(x-5)^2\implies y=\cfrac{1}{4}(x-5)^2[/tex]
