Question 878130
name direction vertex, Axis of symmetry, focus, directrix of the parabola 
x-1=1/4(y-8)^2 
(y-8)^2=4(x-1)
This is an equation of a parabola that opens rightward.
Its basic form: (y-k)^2=4p(x-k), (h,k)=coordinates of the vertex
For given parabola:
vertex: (1,8)
axis of symmetry: y=8
4p=4
p=1
focus(2,8)
directrix: x=0, or y-axis
see graph below:
y=±(4x-4)^.5+8
{{{ graph( 300, 300, -10, 10, -1, 20,(4x-4)^.5+8,-(4x-4)^.5+8) }}}