Question 448876
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Hi
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
x^2+10-4y +1= 0
  4y = x^2 + 11
   y = (1/4)x^2 + 11/4  Vertex (0,11/4)

Standard form is {{{(x -h)^2 = 4p(y -k)}}}, where  the focus is (h,k + p) 
  4y -11 = x^2
  x^2 = 4(y-11/4)  4p = 4, p = 1  focus (0,11/4+1) or (0,15/4)
{{{drawing(300,300,   -6, 6, -6, 6,  grid(1),
circle(0, 11/4,0.3),
circle(0, 15/4,0.3),
graph( 300, 300, -6, 6, -6, 6,0,.25x^2 + 11/4))}}}