Answer
$(x-3)^{2}=-8y$
The latus rectum has endpoints $(-1,-2)$ and $(7,-2).$
Work Step by Step
The focus lies below the vertex, on the line $x=3$, so the parabola opens down.
By table 2,
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h,k-a)}&{y=k+a}&{(x-h)^{2}=-4a(y-k)}& down \\ \hline \end{array}$
$(h,k)=(3,0)$
Given the focus is at $(3,-2)=(h,k-a)$, we find $a:$
$-2=k-a$
$a=2.$
The equation is $\quad (x-3)^{2}=-4(2)(y-0),$
that is, $\quad \fbox{$(x-3)^{2}=-8y$},$
For the latus rectum (line segment parallel to the directrix, containing the focus),
set $y=-2$
$(x-3)^{2}=-8(-2)$
$(x-3)^{2}=16$
$x-3=\pm 4$
$x=3\pm 4$
The latus rectum has endpoints $(-1,-2)$ and $(7,-2).$
The directrix has equation $\quad y=k+a\quad \Rightarrow\quad y=2$