Answer
$(x+4)^2=12(y-1)$
Latus rectum points: $(-10,4),(2,4)$
See graph
Work Step by Step
We are given the parabola:
Focus: $(-4,4)$
Directrix: $y=-2$
Because the directrix is in the form $y=a$, the parabola is vertical. Its standard equation is:
$(x-h)^2=4p(y-k)$
Use the coordinates of the focus and the directrix to determine $h,k,p$:
$(h,k+p)=(-4,4)$
$y=k-p=-2$
$\begin{cases}
h=-4\\
k+p=4\\
k-p=-2
\end{cases}$
$\begin{cases}
h=-4\\
k+p=4\\
k-p+k+p=-2+4
\end{cases}$
$\begin{cases}
h=-4\\
k+p=4\\
2k=2
\end{cases}$
$\begin{cases}
h=-4\\
k+p=4\\
k=1
\end{cases}$
$k+p=4$
$1+p=4$
$p=3$
Determine the parabola's equation:
$(x+4)^2=4(3)(y-3)$
$(x+4)^2=12(y-1)$
Determine the two points defining the latus rectum:
$y=4$
$(x+4)^2=12(4-1)$
$(x+4)^2=36$
$x+4=\pm 6$
$x+4=-6\Rightarrow x_1=-10$
$x+4=6\Rightarrow x_2=2$
$\Rightarrow (-10,4),(2,4)$
Plot the points and graph the parabola:
