[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}\\\\
-------------------------------\\\\
x^2-12y=0\implies x^2=12y\implies (x-0)^2=12(y-0)
\\\\\\
(x-\stackrel{h}{0})^2=4(\stackrel{p}{3})(y-\stackrel{k}{0})[/tex]
so, first off, notice, the squared variable is the "x", meaning is a vertical parabola.
the coefficient of the x² is a positive value, thus the vertical parabola is opening upwards.
the vertex is at 0,0, namely the origin.
the "p" distance is 3.
so the focus point is 3 units above the vertex, and you surely know where that is.
