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

Say you have three known points `p`_{0}, `p`_{1}, and `p`_{2}. 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:

`z`_{0}=a_{0}x^{2}+b_{0}x+c_{0}

z_{1}=a_{1}x^{2}+b_{1}x+c_{1}

Find the quadratic equation that is the difference of the two, and set it equal to zero:

`0=(a`_{0}-a_{1})x^{2}+(b_{0}-b_{1})x+(c_{0}-c_{1})

Use the quadratic formula to find the zeros:

`x=(-(b`_{0}-b_{1})±√((b_{0}-b_{1})^{2}-4(a_{0}-a_{1})(c_{0}-c_{1})))/2(a_{0}-a_{1})

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=h^{2}/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_{x}^{2}/2(F_{y}-d)+F_{y}-(F_{y}-d)/2[/tt]

So you could *technically* write out the equation of a parabola as

`y = x`^{2}/2(F_{y}-d) + -F_{x}x/(F_{y}-d) + F_{x}^{2}/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_{x}^{2}/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.

`a`_{2}((-(b_{0}-b_{1})±√((b_{0}-b_{1})^{2}-4(a_{0}-a_{1})(c_{0}-c_{1})))/2(a_{0}-a_{1}))^{2}+b_{2}((-(b_{0}-b_{1})±√((b_{0}-b_{1})^{2}-4(a_{0}-a_{1})(c_{0}-c_{1})))/2(a_{0}-a_{1}))+c_{2}=a_{0}((-(b_{0}-b_{1})±√((b_{0}-b_{1})^{2}-4(a_{0}-a_{1})(c_{0}-c_{1})))/2(a_{0}-a_{1}))^{2}+b_{0}((-(b_{0}-b_{1})±√((b_{0}-b_{1})^{2}-4(a_{0}-a_{1})(c_{0}-c_{1})))/2(a_{0}-a_{1}))+c_{0}

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.