Answer
$(y+2)^2=4(x+1)$
Latus rectum points: $(0,-4),(0,0)$
See graph
Work Step by Step
We are given the parabola:
Focus: $(0,-2)$
Vertex: $(-1,-2)$
Because the vertex and the focus have the same $y$-coordinate, the parabola is horizontal. Its standard equation is:
$(y-k)^2=4p(x-h)$
Use the coordinates of the vertex to determine $h,k$:
$(h,k)=(-1,-2)$
$h=-1$
$k=-2$
Determine $p$ using the coordinates of the focus:
$(h+p,k)=(0,-2)$
$(-1+p,-2)=(0,-2)$
$-1+p=0$
$p=1$
Determine the parabola's equation:
$(y+2)^2=4(1)(x+1)$
$(y+2)^2=4(x+1)$
Determine the two points defining the latus rectum:
$x=0$
$(y+2)^2=4(0+1)$
$(y+2)^2=4$
$y+2=\pm 2$
$y+2=-2\Rightarrow y_1=-4$
$y+2=2\Rightarrow y_2=0$
$\Rightarrow (0,-4),(0,0)$
Plot the points and graph the parabola:
