Answer
$(x+3)^2=4(y-3)$, $(-5,4),(-1,4)$.
See graph.
Work Step by Step
1. Given focus $(-3,4)$ and directrix $y=2$ (opens upward), we have $p=(4-2)/2=1$, vertex $(-3,3)$, and equation (as in a standard form) $(x+3)^2=4(1)(y-3)$ or $(x+3)^2=4(y-3)$
2. Let $y=4$, we have $(x+3)^2=4(4-3)$ and $x=-3\pm2$, thus we have the two points that define the latus rectum $(-5,4),(-1,4)$
3. See graph.