@Wizlem
You got something wrong there. For t=1 you get (x,y)=(21,10) which should be on this curve, and when you plug in x=21,y=10 into your equation, you end up with -1.1M instead of 0.
I followed up on my own suggestion, as you don't have to deal with square roots (and is easier to avoid mistakes like above ), and it's mostly about solving a quite simple set of linear equations. I noted before that one param will even be free (6 unknowns, 5 equations) like an oddity - I failed to see that this is required, all zeros is always a solution, so it'd be bad if we had a unique solution, as that would be it. And for a conic, one of A,B,C needs to be non-zero. Furthermore, if you have some equation, you can multiply it by any non-zero real to get another set of valid coeffs, so it's a further reason that we need a free param.
Anyway, I get 5X^2 + xy + (1/20)y^2 -220x + 220y = 0 (if you multiply by 20, you can get all-integer coeffs, but I like it this way...)
Since B^2 - 4AC = 0, this turns out to be a parabola.