Answer
$\quad (x+3)^{2}=4(y-3)$
The latus rectum has endpoints $(-1,4)$ and $(-5,4).$
Work Step by Step
1.$\quad $The axis of symmetry is perpendicular to the directrix (it is vertical).
The vertical line through the focus is $x=-3.$
The focus is 2 units above the directrix $\quad \Rightarrow\quad $the parabola opens up.
2.$\quad $By table 2,
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h,k+a)}&{y=k-a}&{(x-h)^{2}=4a(y-k)}&up\\ \hline \end{array}$
The vertex is halfway between the focus and the directrix,
one unit below the focus, one unit above the directrix so $(h,k)=( -3, 3).$
From the directrix $y=k-a=2,$ we find $a$
$3-a=2$
$a=1$
3.$\quad $The equation of the parabola is $\quad (x+3)^{2}=4(1)(y-3)$
that is, $\quad (x+3)^{2}=4(y-3)$
4.$\quad $For the latus rectum (line segment parallel to the directrix, containing the focus),
set $y=4$
$(x+3)^{2}=4(4-3)$
$(x+3)^{2}=4$
$x+3=\pm 2$
$x=-3\pm 2$
The latus rectum has endpoints $(-1,4)$ and $(-5,4).$
