Answer
$(x-2)^2=-8(y+3)$
Latus rectum points: $(-2,-5),(6,-5)$
See graph
Work Step by Step
We are given the parabola:
Focus: $(2,-5)$
Vertex: $(2,-3)$
Because the vertex and the focus have the same $x$-coordinate, the parabola is vertical. Its standard equation is:
$(x-h)^2=4p(y-k)$
Use the coordinates of the vertex to determine $h,k$:
$(h,k)=(2,-3)$
$h=2$
$k=-3$
Determine $p$ using the coordinates of the focus:
$(h,k+p)=(2,-5)$
$(2,-3+p)=(2,-5)$
$-3+p=-5$
$p=-2$
Determine the parabola's equation:
$(x-2)^2=4(-2)(y-(-3))$
$(x-2)^2=-8(y+3)$
Determine the two points defining the latus rectum:
$y=-5$
$(x-2)^2=-8(-5+3)$
$(x-2)^2=16$
$x-2=\pm 4$
$x-2=-4\Rightarrow x_1=-2$
$x-2=4\Rightarrow x_2=6$
$\Rightarrow (-2,-5),(6,-5)$
Plot the points and graph the parabola:
