A 3rd degree polynomial is uniquely defined by 4 points GIVEN 4 parameters, 1 for each point !!!!!!

Like this :

If P(t1)=P1, P(t2)=P2, P(t3)=P3, P(t4)=P4

then
(1) P(t) = P1*[(t-t2)*(t-t3)*(t-t4)]/[(t1-t2)*(t1-t3)*(t1-t4)] + P2*... rotate the variables ... + P3*... + P4*...

This shows that you need 16 reals to define a unique Cubic ! A cubic defined as a polynomial and not a shape !

Now what happens if you dont specify t1, ... t4 and add 2 more unknowns t5, t6 and 2*3=6 constraints P(t5)=P5 and P(t6)=P6

Then I have 18 unknowns( t1,..,t6 and 3*4=12 polynomial coeffs ) and 6*3=18 equations ! That should be all right !

Currently I dont know if there is a solution but if there is one, there is only one !

I have to solve that 18*18 non linear shit ! LOL
In fact I use (1) thus I eliminate all the 12 coeffs and 12 equations !!! I only have 6 equations and 6 unknowns (t1,...,t6)

P(t5)=P5
P(t6)=P6

Then I let you solve :] LOL

Charles B