Answer
$(y+2)^{2}=-8(x+1)$
The latus rectum has endpoints $(-3,-6)$ and $(-3,2).$
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, $y= -2.$
The focus is $4$ units left of the directrix, so it opens left.
The vertex is halfway between the focus and directrix,
$a=2$ units from either, $(h,k)=(-1,-2).$
2.$\quad $ Using table 2:
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h-a,k)}&{x=h+a}&{(y-k)^{2}=-4a(x-h)} &left\\ \hline \end{array}$
The equation is $(y+2)^{2}=-4(2)(x+1)$
that is, $\quad (y+2)^{2}=-8(x+1)$
3.$\quad $For the latus rectum (line segment parallel to the directrix, containing the focus),
set $x=-3$
$(y+2)^{2}=-8(-3+1)$
$(y+2)^{2}=16$
$y+2=\pm 4$
$y=-2\pm 4$
The latus rectum has endpoints $(-3,-6)$ and $(-3,2).$
