Answer
$(y-4)^2=12(x+1)$
Latus rectum points: $(2,-2),(2,10)$
See graph
Work Step by Step
We are given the parabola:
Focus: $(2,4)$
Directrix: $x=-4$
Because the directrix is in the form $x=a$, the parabola is horizontal. Its standard equation is:
$(y-k)^2=4p(x-h)$
Use the coordinates of the focus and the directrix to determine $h,k,p$:
$(h+p,k)=(2,4)$
$x=h-p=-4$
$\begin{cases}
h+p=2\\
k=4\\
h-p=-4
\end{cases}$
$\begin{cases}
h+p=2\\
k=4\\
h-p+h+p=-4+2
\end{cases}$
$\begin{cases}
h+p=2\\
k=4\\
2h=-2
\end{cases}$
$\begin{cases}
h+p=2\\
k=4\\
h=-1
\end{cases}$
$h+p=2$
$-1+p=2$
$p=3$
Determine the parabola's equation:
$(y+2)^2=4(1)(x+1)$
$(y-4)^2=4(3)(x+1)$
$(y-4)^2=12(x+1)$
Determine the two points defining the latus rectum:
$x=2$
$(y-4)^2=12(2+1)$
$(y-4)^2=36$
$y-4=\pm 6$
$y-4=-6\Rightarrow y_1=-2$
$y-4=6\Rightarrow y_2=10$
$\Rightarrow (2,-2),(2,10)$
Plot the points and graph the parabola:
