Question 859895
this is where You are going...
the vertex form of a Parabola opening up(a>0) or down(a<0), {{{y=a(x-h)^2 +k}}} 
where(h,k) is the vertex  
x^2 - 8x - 16y + 16 = 0
(x-4)^2 - 16  -16y + 16 = 0
{{{y = (1/16)(x-4)^2}}} V = (4,0) x = 4 line of symmetry 
a = 1/4p = 1/16, p = 4
focus is (h,k + p) (4,4)
With Directrix y = (k - p) y = -4
{{{drawing(300,300,   -6, 6, -6, 6, grid(1), blue(line(4,6,4,-6)) ,
circle(4, 4,0.2),
circle(4, 0,0.2),
graph( 300, 300, -6, 6, -6, 6,0,-4, (1/16)(x-4)^2) )}}}