Answer
$x^2=-12y$
Latus rectum points: $(-6,-3),(6,-3)$
See graph
Work Step by Step
We are given the parabola:
Focus: $(0,-3)$
Vertex: $(0,0)$
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 $h,k$:
$(h,k)=(0,0)$
$h=0$
$k=0$
Determine $p$ using the coordinates of the focus:
$(h,k+p)=(0,-3)$
$(0,0+p)=(0,-3)$
$p=-3$
Determine the parabola's equation:
$(x-0)^2=4(-3)(y-0)$
$x^2=-12y$
Determine the two points defining the latus rectum:
$y=-3$
$x^2=-12(-3)$
$x^2=36$
$x=\pm 6$
$\Rightarrow (-6,-3),(6,-3)$
Plot the points and graph the parabola:
$x^2=-12y$
Latus rectum points: $(-6,-3),(6,-3)$
See graph
