Answer
$${\text{Click to see the graph}}$$
Work Step by Step
$$\eqalign{
& 8y = - 3{x^2} \cr
& {x^2} = - \frac{8}{3}y \cr
& {\text{The equation is in the form }}{x^2} = 4py \cr
& {x^2} = - \frac{8}{3}y:{\text{ }}{x^2} = 4py \cr
& - \frac{8}{3}y = 4py \cr
& p = - \frac{2}{3} \cr
& p < 0 \cr
& {\text{The parabola opens downward}} \cr
& {\text{The focus of a parabola of the form }}{x^2} = 4py{\text{ is}} \cr
& {\text{Focus }}\left( {0,p} \right) \Rightarrow {\text{Focus }}\left( {0, - \frac{2}{3}} \right) \cr
& {\text{The directrix of a parabola of the form }}{x^2} = 4py{\text{ is}} \cr
& y = - p \Rightarrow y = \frac{2}{3} \cr
& \cr
& {\text{Graph}} \cr} $$
