Converting a function of $t$ into a function of $y$

Question

I have an equation that looks like this:

$$ y = \sqrt{ ((at^3+bt^2+ct+d) - (et+f))^2 + ((gt^3+ht^2+it+j) - (kt+l))^2 } $$ [original image: https://i.stack.imgur.com/oSw67.png]

But I have to make this a function of $y$ instead of a function of $t$, and I'm totally stuck on where to even begin. Could anyone give me advice on where to start, or perhaps solve my problem?

EDIT: (Context)
One circle moves along a cubic bezier curve and another moves along a linear bezier curve. Between $0 ≤ t ≤ 1$, at what t values does the two circles collide if the radius of them both is a total of $y$ (distance between the two circles)?

Answer 1

So for a fixed, known $y$, you want to solve $$f(t)=g(t) - y^2 = 0$$ for $t\in [0,1]$, where $g$ (and hence $f$) are sixth-degree polynomials in $t$. Generally speaking, it won't be possible to solve for this $t$ exactly. Standard practice in computer graphics is to find $t$ numerically using a polynomial root-finding algorithm: my personal recommendation is the Jenkins-Traub algorithm (source code here: www.crbond.com/download/misc/rpoly.cpp‎), which is both fast and robust.

EDIT: To use rpoly, simplify the above polynomial so that it has the form $$a_0 t^6 + a_1t^5 + a_2 t^4 + a_3 t^3 + a_4 t^2 + a_5 t + a_6 = 0.$$

I am going to assume that $a_0$ is well away from $0$ (otherwise there are stability complications beyond the scope of this answer).

You can find all of the zeros using code along the following lines:

double op[7];
double zeror[7];
double zeroi[7];
op[0] = a0;
op[1] = a1;
op[2] = a2;
op[3] = a3;
op[4] = a4;
op[5] = a5;
op[6] = a6;

rpoly(op, 6, zeror, zeroi, NULL);

The real and imaginary parts of the roots will now be inside zeror and zeroi. Ignore all non-real roots (those whose imaginary part is greater than some threshold) and look for those real roots that lie between 0 and 1. I'm not sure if you're looking for the first time the two balls collide, or all times they collide, etc. but whatever case it is you should be able to extract that information from the roots.