### Author Topic: Three intersecting parabolas  (Read 3393 times)

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#### ZippyDee

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• Why not zoidberg? ##### Three intersecting parabolas
« on: May 03, 2012, 08:01:19 pm »

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.
« Last Edit: May 03, 2012, 10:52:08 pm by ZippyDee »

Pushpins 'n' stuff... #### Xeda112358

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• Calc-u-lator, do doo doo do do do. ##### Re: Three intersecting parabolas
« Reply #1 on: May 03, 2012, 08:43:15 pm »
Hmm, wait, so I need to clarify.

Do all the parabolas have the same directrix?
When you say "assume that the three foci do NOT all lie on a single line parallel to the directrix," are you just saying make sure all the foci don't lie on the same line?

#### ZippyDee

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• Why not zoidberg? ##### Re: Three intersecting parabolas
« Reply #2 on: May 03, 2012, 10:17:40 pm »
Yes, they all have the same directrix. The foci can all lie on the same line, as long as that line is not parallel to the directrix.

Pushpins 'n' stuff... #### Xeda112358

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• Calc-u-lator, do doo doo do do do. ##### Re: Three intersecting parabolas
« Reply #3 on: May 03, 2012, 10:57:18 pm »
But the directrix is constant for parabolas, so you are saying as long as the slope is non-constant?

#### ZippyDee

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• Why not zoidberg? ##### Re: Three intersecting parabolas
« Reply #4 on: May 03, 2012, 10:59:27 pm »
If the foci are all at the same y-coordinate, they will never intersect no matter what the value of d is. As long as their y-coordinates are not all the same, it is solvable. 