Answer
$(y+2)^2=8(x-4)$
Latus rectum points: $(6,-6),(6,2)$
See graph
Work Step by Step
We are given the parabola:
Focus: $(6,-2)$
Vertex: $(4,-2)$
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 $h,k$:
$(h,k)=(4,-2)$
$h=4$
$k=-2$
Determine $p$ using the coordinates of the focus:
$(h+p,k)=(6,-2)$
$(4+p,-2)=(6,-2)$
$4+p=6$
$p=2$
Determine the parabola's equation:
$(y+2)^2=4(2)(x-4)$
$(y+2)^2=8(x-4)$
Determine the two points defining the latus rectum:
$x=6$
$(y+2)^2=8(6-4)$
$(y+2)^2=16$
$y+2=\pm 4$
$y+2=-4\Rightarrow y_1=-6$
$y+2=4\Rightarrow y_2=2$
$\Rightarrow (6,-6),(6,2)$
Plot the points and graph the parabola:
