Intersections of two parabolas given focii

Question

As part of Voronoi's algorithm, I need to calculate the intersection of two parabolas to compute a breakpoint at run time.

I've spent literally 8 hours on this, and I've only gotten my equations to cancel out. I've got the foci of both parabolas (pointA, pointB) and the line of their directrix (line).

I would think it'd be easy enough to solve for using $x^2 = 4py$, but it isn't. I tried to figure out an alternative algorithm of computing the circle from pointA, pointB, and the line, such that it's center would be the break point. But all of my attempts either resulted in unsolvable equations or ones that just cancelled out.

I know what the equations are for circles & parabolas, where am I going wrong?

Answer 1

You have said in a comment that each directrix is parallel to the $x$-axis. Let's suppose the foci are the points $(f_{x1}, f_{y1})$ and $(f_{x2}, f_{y2})$ while the directrices are the lines $y=d_1$ and $y=d_2$ respectively.

Then the parabolae are $y=\frac12\dfrac{(x-f_{x1})^2}{f_{y1}-d_1}+\frac12f_{y1}+\frac12d_1$ and $y=\frac12\dfrac{(x-f_{x2})^2}{f_{y2}-d_2}+\frac12 f_{y2}+\frac12 d_2$

and you need to solve $(f_{y2}-d_2)\left((x-f_{x1})^2+f_{y1}^2-d_1^2\right) = (f_{y1}-d_1)\left((x-f_{x2})^2+f_{y2}^2-d_2^2\right)$ to find the $x$ coordinates of the intersections. This is a quadratic in $x$, possibly degenerate if $f_{y1}-d_1=f_{y2}-d_2$; as ever with solving a quadratic, you should expect two, one or zero real solutions.

Answer 2

Assume the parabolas are:

$$(y-y_1)^2=4a(x-x_1);(y-y_2)^2=4b(x-x_2)$$

Now parametrize those parabolas by:

$$(x_1+at^2,y_1+2at);(x_2+bT^2,y_2+2bT)$$

Since you're looking for intersection: $$x_1+at^2=x_2+aT^2;y_1+2at=y_2+2bT\\T=\frac{y_1+2at-y_2}{2b};T^2=\frac{x_1+at^2-x_2}{a}$$

SO: $$\frac{x_1+at^2-x_2}{a}=\left(\frac{y_1+2at-y_2}{2b}\right)^2$$

Let $X=x_1-x_2,Y=y_1-y_2$ and then solve the quadratic: $$4b^2(X+at^2)=a(4at^2+Y^2-4Yat)\\4a(b^2-a)t^2+4a^2Yt+(4b^2X-aY^2)=0$$

Now $t$ would be something like:

$$t=\frac{-4a^2Y\pm\sqrt{16a^4Y^2-16a(b^2-a)(4b^2X-aY^2)}}{8a(b^2-a)}$$

Put back to get Coordinates.