Answer
$(y+2)^2=4(x+1)$, $(0,-4),(0,0)$.
See graph.
Work Step by Step
1. Given vertex $(-1,-2)$ and focus $(0,-2)$ (opens rightward), we have $p=2$ and equation (as in a standard form) $(y+2)^2=4(1)(x+1)$ or $(y+2)^2=4(x+1)$
2. Let $x=0$, we have $(y+2)^2=4(0+1)$ and $y=-2\pm2$, thus we have the two points that define the latus rectum $(0,-4),(0,0)$
3. See graph.