Question 1028982
 {{{x^2-2x-36y+1=0}}}
Complete the square for the x if needed; but it is not needed.
{{{x^2-2x+1=36y}}}
{{{(x-1)^2=36y}}}
While not yet in standard form, this is in a form which would result IF you had derived the equation using the directrix and the focus, neither of which are given.
This is a parabola with a vertex minimum, opening upward.  


If p is how far the vertex is from the focus and the directrix, then
{{{4p=36}}}
{{{p=9}}}
and 
The vertex is at (1,0).
Focus (10,0)
Directrix y=-8
-
Standard form equation is  {{{y=(1/36)(x-1)^2+0}}}
{{{graph(350,350,-6,6,-6,6,(1/36)(x-1)^2)}}}





{{{(x-3)^2=-12(y-4)}}}----
This parabola opens downward and has a vertex maximum.  Again, the equation is in a manner which could come from deriving if you had been given the focus and directrix.
Read the vertex directly from this equation, since you can.
Vertex  ((3,4).

I will let you figure out the directrix and focus.