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