Question 195651
Keep in mind that the parabola form we'll use is 


{{{(x-h)^2=4p(y-k)}}} where "p" is the distance from the vertex to the focus.



It turns out that the distance from the directrix to the vertex is equal to the distance from the vertex to the focus. Since the distance from the directrix y = 2 to the vertex (0,0) is 2 units, this means that {{{p=2}}}



Also, since the vertex is (0,0), this means that {{{h=0}}} and {{{k=0}}}



So all we need to do is plug these values in and isolate "y":



{{{(x-h)^2=4p(y-k)}}} Start with the given equation.



{{{(x-0)^2=4(2)(y-0)}}} Plug in {{{p=2}}}, {{{h=0}}} and {{{k=0}}}



{{{x^2=4(2)y}}} Simplify



{{{x^2=8y}}} Multiply



{{{(1/8)x^2=y}}} Multiply both sides by {{{1/8}}} to isolate "y".



{{{y=(1/8)x^2}}} Rearrange the terms.



Now because the parabola has the directrix above the vertex, this means that the parabola is opening down. So this means we need to attach a negative sign to the coefficient: {{{y=-(1/8)x^2}}}




So the equation is {{{y=-(1/8)x^2}}}