Answer
$(x+4)^2=12(y-1)$, $(-10,4),(2,4)$.
See graph.
Work Step by Step
1. Given focus $(-4,4)$ and directrix $y=-2$ (opens upward), we have $p=(4+2)/2=3$, vertex $(-4,1)$, and equation (as in a standard form) $(x+4)^2=4(3)(y-1)$ or $(x+4)^2=12(y-1)$
2. Let $y=4$, we have $(x+4)^2=12(4-1)$ and $x=-4\pm6$, thus we have the two points that define the latus rectum $(-10,4),(2,4)$
3. See graph.