Answer
$y^2=16x$
Latus rectum points: $(4,-8),(4,8)$
See graph below.
Work Step by Step
We are given the parabola with:
Focus: $(4,0)$
Vertex: $(0,0)$
Because the vertex and the focus have the same $y$-coordinate, the parabola is horizontal (opens to the left/right). 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, which is $(x+p, k)$:
$(h+p,k)=(4,0)$
$(0+p,0)=(4,0)$
$p=4$
Substituting the values of $h, k, \text{ and } p$ in the standard equation of a horizontal parabola gives:
$(y-0)^2=4(4)(x-0)$
$y^2=16x$
Since the vertex is at the origin and $p=4$, the coordinates of the latus rectum can be determined by letting $x=4$ then solvig for $y$:
$y^2=16x$
$y^2=16(4)$
$y^2=64$
$y=\pm 8$
$y=-8\Rightarrow y_1=-8$
$y=8\Rightarrow y_2=8$
$\Rightarrow (4,-8),(4,8)$
Plot the vertex and the latus rrectum endpoints, then connect them using a smooth curve.
