Question 861647
You have two choices.  One is to use the basic developed formulaic knowledge about a parabola.  The other is to use the distance formula to derive the equation which you want.  


There is a general point on the parabola, (x, y).  There is the line y=-2 and there is the focus (0,2).  The variable point for the directrix line is (x,-2).  Can you begin to draw this and see that the vertex will be (0,0) the origin?  Continue to draw a representation of this parabola, since it will also help you in the use of the distance formula.


Distance parabola to focus = Distance parabola to directrix


{{{sqrt((x-0)^2+(y-2)^2)=sqrt((x-x)^2+(y-(-2))^2)}}}; Does this equation make sense to you?  When it makes sense, then simplify it, and work it into solved as y in terms of x.

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{{{sqrt(x^2+(y-2)^2)=sqrt(0+(y+2)^2)}}}
{{{x^2+(y-2)^2=(y+2)^2}}}
{{{x^2+y^2-4y+4=y^2+4y+4}}}
{{{x^2-4y=4y}}}-----this was a combination step, 
{{{4y=x^2-4y}}}------symmetric property
{{{8y=x^2}}}
{{{highlight(y=(1/8)x^2)}}}.