Answer
$y^{2}=16x$
The latus rectum has endpoints $(4,8)$ and $(4,-8)$
Work Step by Step
Given: $a=4,\ (h,k)=(0,0)$
1. The focus and the vertex lie on the horizontal line $\quad y=0.$
2. The focus $(4,0)$ is right of the vertex $(0,0)$
So the parabola opens to the right.
3. By table 2, the equation is$\quad (y-k)^{2}=4a(x-h)$
that is, $\quad y^{2}=4(4)x,$
$y^{2}=16x$
4. For $x=4, \quad y^{2}=64\Rightarrow y=\pm 8$
The latus rectum (line segment parallel to the axis, containing the focus)
has endpoints $(4,8)$ and $(4,-8)$
5. Directrix: $\quad x=h-a\quad \Rightarrow\quad x=-4$
We have enough details for the graph.
