check the picture below.
so the focus point is at 5, -1 and the directrix is above it, meaning is a vertical parabola, and is opening downwards, like in the picture.
keep in mind that the vertex is half-way between those two fellows, at a "p" distance from either, in this case 1 unit, since the parabola is opening downwards, "p" is negative then, or -1, and the vertex will be from 5, -1 up one unit, so 5,0.
[tex]\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}})
\\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})}
\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 (x-5)^2=4(-1)(y-0)\implies (x-5)^2=-4y
\\\\\\
-\cfrac{1}{4}(x-5)^2=y[/tex]
