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 (zn)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=(Fy-d)/2
h=Fx
k=Fy-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(Fy-d)
b=-Fx/(Fy-d)
c=Fx2/2(Fy-d)+Fy-(Fy-d)/2[/tt]
So you could technically write out the equation of a parabola as
y = x2/2(Fy-d) + -Fxx/(Fy-d) + Fx2/2(Fy-d)+Fy-(Fy-d)/2
As you can see, it gets pretty hectic pretty quickly. For example, the quadratic formula would then be:
x=(Fx/(Fy-d)±√((-Fx/(Fy-d))2-(2/(Fy-d))(Fx2/2(Fy-d)+Fy-(Fy-d)/2)))/(1/(Fy-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 Fnx and Fny for each variable of the corresponding n value.
