Answer
$y^2=-16x$
Latus rectum points: $(-4,-8),(-4,8)$
See graph
Work Step by Step
We are given the parabola:
Focus: $(-4,0)$
Vertex: $(0,0)$
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 $h,k$:
$(h,k)=(0,0)$
$h=0$
$k=0$
Determine $p$ using the coordinates of the focus:
$(h+p,k)=(-4,0)$
$(0+p,0)=(-4,0)$
$p=-4$
Determine the parabola's equation:
$(y-0)^2=4(-4)(x-0)$
$y^2=-16x$
Determine the two points defining the latus rectum:
$x=-4$
$y^2=-16(-4)$
$y^2=64$
$y=\pm 8$
$y=-8\Rightarrow y_1=-8$
$y=8\Rightarrow y_2=8$
$\Rightarrow (-4,-8),(-4,8)$
Plot the points and graph the parabola:
