Three intersecting parabolas

I've been thinking about this all day and I can't figure it out...

Say you have three known points p₀, p₁, and p₂. Each of these points serves as the focus of a parabola, and all three parabolas share a single line at y=d as their directrix. Assume that d is such that all three parabolas open in the same direction, and assume that the three foci do NOT all lie on a single line parallel to the directrix.

So the idea is to solve for value of d that makes all three parabolas intersect at a single point. There should only be one value that makes this true.

Basically, so far I've got quite a bit conceptually figured out in terms of what needs to happen...but I don't really know how to make it happen.

Here's kinda what I have:

The intersection point(s) of two parabolas (z_n)can be calculated fairly easily:
z₀=a₀x²+b₀x+c₀ z₁=a₁x²+b₁x+c₁
Find the quadratic equation that is the difference of the two, and set it equal to zero:
0=(a₀-a₁)x²+(b₀-b₁)x+(c₀-c₁)
Use the quadratic formula to find the zeros:
x=(-(b₀-b₁)±√((b₀-b₁)²-4(a₀-a₁)(c₀-c₁)))/2(a₀-a₁)

A parabola can also be written as (x-h)²=4p(y-k), where (h,k) is the vertex of the parabola, (h,k+p) is the focus, and the directrix lies at y=k-p
With a focal point F and a directrix at y=d, all other necessary variables can be calculated as follows:
p=(F_y-d)/2 h=F_x k=F_y-p a=1/4p b=-h/2p c=h²/4p+k

Through this, all variables can be simplified down to only use F and d:
a=1/2(F_y-d)
b=-F_x/(F_y-d)
c=F_x²/2(F_y-d)+F_y-(F_y-d)/2[/tt]

So you could technically write out the equation of a parabola as
y = x²/2(F_y-d) + -F_xx/(F_y-d) + F_x²/2(F_y-d)+F_y-(F_y-d)/2

As you can see, it gets pretty hectic pretty quickly. For example, the quadratic formula would then be:
x=(F_x/(F_y-d)±√((-F_x/(F_y-d))²-(2/(F_y-d))(F_x²/2(F_y-d)+F_y-(F_y-d)/2)))/(1/(F_y-d))

Remember, this needs to be solving for d...

Really, I don't know where to go from here. Anyone have any insight?

---EDIT:
Okay, so I'm slowly figuring things out.

The equation is crazy complex, but you have to solve for the intercepts of two parabolas, then plug that quadratic formula calculation in for X in one of those equations, and set it equal to the equation for the third parabola with the quadratic formula calculation plugged in for X as well.

a₂((-(b₀-b₁)±√((b₀-b₁)²-4(a₀-a₁)(c₀-c₁)))/2(a₀-a₁))²+b₂((-(b₀-b₁)±√((b₀-b₁)²-4(a₀-a₁)(c₀-c₁)))/2(a₀-a₁))+c₂=a₀((-(b₀-b₁)±√((b₀-b₁)²-4(a₀-a₁)(c₀-c₁)))/2(a₀-a₁))²+b₀((-(b₀-b₁)±√((b₀-b₁)²-4(a₀-a₁)(c₀-c₁)))/2(a₀-a₁))+c₀

Looks like fun, right? Now substitute the As, Bs, and Cs with their F/d equivalents that I mentioned earlier, accounting for the fact that the Fs have to be F_nx and F_ny for each variable of the corresponding n value.