Answer
$(y+2)^2=8(x-4)$, $(6,-6),(6,2)$.
See graph.
Work Step by Step
1. Given vertex $(4,-2)$ and focus $(6,-2)$ (opens to the right), we have $p=6-4=2$ and equation (as in a standard form) $(y+2)^2=4(2)(x-4)$ or $(y+2)^2=8(x-4)$
2. Let $x=6$, we have $(y+2)^2=8(6-4)$ and $y=-2\pm4$, thus we have the two points that define the latus rectum $(6,-6),(6,2)$
3. See graph.