Find the equation of the parabola in which the ends of the latus rectum Find the equation of the parabola in which the ends of the latus rectum have the coordinates $(-1,5)$ and $(-1,-11)$ and the vertex is $(-5,-3)$.

I could think of assuming the equation of parabola as $(y-k)^2=4a(x-h)$ and plug in those three points to get three linear equations and solve for the unknowns. But, I wonder if there's any other approach?
$h$ and $k$ are the coordinates of the vertex, hence $h=-5$ and $k=-3$. As focus is $F=(-1,-3)$ (midpoint of latus rectum) and vertex $V=(-5,-3)$, then $a=VF=4$.
Vertex $(h,k) = (-5,-3)$.

We know, Length of Latus rectum $=4a$, and the end points are $(-1,5)$ and $(-1,-11)$.

Now using the distance formula, we have
$$4a=[ (-1+1)^2 + (-11-5 )^2 ] ^{1/2}$$

or,
$$4a =(0+256)^{1/2}$$

or $4a=16$. Therefore $a=4$.

So the required equation of parabola is 
$$(y-k)^2 = 4a(x-h)$$

or,
$$(y+3)^2= 4\cdot4(x+5)$$

or,
$$y^2+ 6y +9 = 16x + 80$$

or
$$y^2 + 6y - 16x - 71 = 0$$