check the picture below
so... it looks more or less like so
now, notice the "p" distance, bear in mind that, if the parabola opens downwards, "p" is a negative unit, so, in this case -4
thus
[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}[/tex]
[tex]\bf -------------------------------\\\\
(x-{{ h}})^2=4{{ p}}(y-{{ k}})\quad
\begin{cases}
h=0\\
k=0\\
p=-4
\end{cases}\implies (x-0)^2=4(-4)(y-0)
\\\\\\
x^2=-16y\implies \cfrac{x^2}{-16}=y\implies -\cfrac{1}{16}x^2=y[/tex]
