Question 448829
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Hi
write an equation for a graph that is the set of all points in the plane that
 are equidistant from the given point(0,2) and y = -2
Parabola opening upward:  C(0,0)
 4py = x^2    p = 2 
  8y = x^2   or y = x^2/8
{{{drawing(300,300,  -10,10,-10,10, grid(1), 
circle(0, 2,0.3),
graph( 300, 300,-10,10,-10,10,0, -2, (1/8)x^2))}}}

Standard Form of an Equation of a Circle is {{{(x-h)^2 + (y-k)^2 = r^2}}} 
where Pt(h,k) is the center and r is the radius

 Standard Form of an Equation of an Ellipse is {{{(x-h)^2/a^2 + (y-k)^2/b^2 = 1 }}}
where Pt(h,k) is the center and a and b  are the respective vertices distances from center.

Standard Form of an Equation of an Hyperbola is  {{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1}}} where Pt(h,k) is a center  with vertices 'a' units right and left of center.
Standard Form of an Equation of an Hyperbola opening up and down is:
  {{{(y-k)^2/b^2 - (x-h)^2/a^2 = 1}}} where Pt(h,k) is a center  with vertices 'b' units up and down from center.

Using the vertex form of a parabola opening up or down, {{{y=a(x-h)^2 +k}}}
 where(h,k) is the vertex
 The standard form is {{{(x -h)^2 = 4p(y -k)}}}, where  the focus is (h,k + p)