so hmmm if you notice the picture below
bear in mind that, the vertex is "p" distance from the focus, and "p" distance from the directrix
now, the focus is at 4,6 and the directrix below that, meaning the parabola is vertical and opening upwards
and the vertex is half-way between the focus point and the directrix
[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}\\\\
-----------------------------\\\\
(x-4)^2=4(2)(y-4)\implies (x-4)^2=8(y-4)
\\\\\\
\cfrac{(x-4)^2}{8}=y-4\implies \cfrac{1}{8}(x-4)^2+4=y[/tex]
