Answer
$${\text{Click to see the graph}}$$
Work Step by Step
$$\eqalign{
& {x^2} = 12y \cr
& {\text{The equation is in the form }}{x^2} = 4py \cr
& 12y = 4py \Rightarrow p = 3 \cr
& p > 0 \cr
& {\text{The parabola opens upward}} \cr
& {\text{The focus of a parabola of the form }}{x^2} = 4py{\text{ is}} \cr
& {\text{Focus }}\left( {0,p} \right) \Rightarrow {\text{Focus }}\left( {0,3} \right) \cr
& {\text{The directrix of a parabola of the form }}{x^2} = 4py{\text{ is}} \cr
& y = - p \Rightarrow y = - 3 \cr
& \cr
& {\text{Graph}} \cr} $$
