Answer
$(y-4)^{2}=12(x+1)$
The latus rectum has endpoints $(2,-2)$ and $(2,10).$
Work Step by Step
1.$\quad $The directrix is vertical, so the parabola opens left or right.
The focus and the vertex lie on a horizontal line, $y=4$.
The focus is 6 units right of the directrix, so it opens right.
The vertex is halfway between the focus and directrix,
3 units from either, $(h,k)=(-1,4),\ \quad a=3.$
2.$\quad $ Using table 2:
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h+a, k)}&{x=h-a}&{(y-k)^{2}=4a(x-h)}& right\\ \hline \end{array}$
The equation is $(y-4)^{2}=4(3)(x+1)$
that is, $\quad (y-4)^{2}=12(x+1)$
3.$\quad $For the latus rectum (line segment parallel to the directrix, containing the focus),
set $x=2$
$(y-4)^{2}=12(2+1)$
$(y-4)^{2}=36$
$y-4=\pm 6$
$y=4\pm 6$
The latus rectum has endpoints $(2,-2)$ and $(2,10).$