Question 1134394
I don't know how to solve your problem. All I can do is write down for you the equation of a parabola.

There are two kinds, vertical and horizontal. IF the parabola is vertical, along y, then it is of the form {{y = Ax^2 + B}}. 

You derive this by recalling that the distance to the focus is equal to the distance to the directrix.

{{(x - a)^2 + (y-b)^2 = (y-k)^2}}

Here the focus is a point of coordinates (a,b). And the directrix has equation y = k.

I think the equation you're looking for is similar to this one, which is why I go through its steps:

{{y^2 - 2ky + k^2 = y^2 -2by + b^2 + (x-a)^2}}

y^2 cancels out, as expected, and we end up with

{{y(2b - 2k) = b^2 - k^2 + (x-a)^2}} or
{{y = (x-a)^2/[2(b-k)] + (b+k)/2}}

The other equation for the parabola, has a physical explanation. It's the trajectory of a point mass, dropped from a plane, or shot through a canon. There is no acceleration in the horizontal direction, so v = constant. 
And thus x = vt.
At the same time, there is acceleration along the y direction, that of gravity. And naturally you have v = gt and {{y = 1/2 g t^2 =1/2 g(x/v)^2 = 1/2 g/v^2 x^2}}

I was hoping someone else solved this problem for you but, if they did, I missed it. Anyway, I hope this helps.