# Is there a simple way to turn a coordinate system “inside out”?

I have the following function, depicted on this image:

The strange thing is, that I only know the coordinate system from the "inside out", that means I know the lengths b, c, e and f, but I don't know a and d. Neither do I know the function. My goal is to calculate a and d. I could assume the function is quadratic, (even if it's not, it would be sufficient as an approximation, if I swapped the two axes).

I also know the length a+d. It's as if the two axes were a thread being stretched over the edge of a cube, and I can move it, and I know all the markings on it, but not the position where it contacts the edge.

The problem is, if I try to use my assumptions on the function for my calculations, I end up with

g(a) = d
g(a+b) = d + e
g(a+b+c) = d + e + f


As I don't know a and d, this results in a big mess in my calculations, and I can't get anywhere, because I don't know the origin of my coordinate system.

As I know the "outside" part of my coordinate system better than the "inside" of it, is there any transformation I can use to make my job easier? (or a completely different strategy to solve my original problem?)

-
Would it help to put the origin at the point you currently call $(a,d)$? The lower left corner becomes $(-a,-d)$ but is decoupled from the other lengths. – Ross Millikan Nov 30 '12 at 22:12

If you know $b,c,e,f$ you have three points on a parabola so should be able to determine it. As in my comment, let the origin be at your $(a,d)$. Then your functional form is $x=py^2+qy$. There is no constant term as it goes through the origin. If we substitute in the known points we get $b=pe^2+qe, (b+c)=p(e+f)^2+q(e+f)$ This is two equations in two unkowns $(p$ and $q)$.
by the way, shouldn't it be $b=pe^2+qe$ ? – vsz Nov 30 '12 at 22:44
If the origin was moved to $(a,d)$ then $g(b) = e$, not $d$ – vsz Nov 30 '12 at 23:33