Answer
$(x-2)^2=-8(y+3)$, $(-2,-5),(6,-5)$.
See graph
Work Step by Step
1. Given vertex $(2,-3)$ and focus $(2,-5)$ (opens down), we have $p=-3+5=2$ and equation (as in a standard form) $(x-2)^2=-4(2)(y+3)$ or $(x-2)^2=-8(y+3)$
2. Let $y=-5$, we have $(x-2)^2=-8(-5+3)=16$ and $x=2\pm4$, thus we have the two points that define the latus rectum $(-2,-5),(6,-5)$
3. See graph.