Answer
$(x+4)^{2}=12(y-1)$
The latus rectum has endpoints $(-10,4)$ and $(2,4).$
Work Step by Step
1.$\quad $The directrix is horizontal, so the parabola opens up or down.
The focus and the vertex lie on a verticall line, $x=-4.$
The focus is $6$ units above the directrix, so it opens up.
The vertex is halfway between the focus and directrix,
$a=3$ units from either, $(h,k)=(-4,1).$
2.$\quad $ Using 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 equation is $(x+4)^{2}=4(3)(y-1)$
that is, $\quad (x+4)^{2}=12(y-1)$
3.$\quad $For the latus rectum (line segment parallel to the directrix, containing the focus),
set $ y=4$
$(x+4)^{2}=12(4-1)$
$(x+4)^{2}=36$
$x+4=\pm 6$
$x=-4\pm 6$
The latus rectum has endpoints $(-10,4)$ and $(2,4).$
