check the picture below, based that the directrix is a horizontal line, and the vertex is below it, we know is a vertical parabola, and is opening downwards.
notice the distance "p" from the vertex to the directrix, is just 1/2
[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}})\qquad
\begin{cases}
h=2\\
k=2\\
p=\frac{1}{2}
\end{cases}\implies (x-2)^2=4\left( \frac{1}{2} \right)(y-2)
\\\\\\
(x-2)^2=2(y-2)\implies \cfrac{1}{2}(x-2)^2=y-2\implies \boxed{\cfrac{1}{2}(x-2)^2+2=y}[/tex]
