Examination of the bezier curve under different conditions (followup from previous post)

I used http://live.sympy.org/ for symbolic computations at this point, another online site that nice provides math analysis tools.

I'd mention as in the previous posting on this subject matter, of supplying a rotation and translation of coordinate where it were assumed the component dx/dt on the parametric curve of the bezier function where neither simultaneously zero at a supplied point, and that we would assumed dy/dt were zero at such point regarding a time t  in the range of [0,1].

The supplied parametric equation on the rotated coordinate and translated coordinate frame for the relative y axis then can be described at such point of time t as


3at(−t+1)2+3bt2(−t+1)

equal to a point value p on the curves interval representing a relative y maximum.  Here points a and b are control points of the third order bezier function, noting that controls points y_0 and y_3 are simultaneously zero on the rotated coordinate axis.  Unlike the previous posting where assumptions were given such that a = b, we provide the possibility of control points a not equal to b in this examination.

The first derivative of this equation is

9at2−12at+3a−9bt2+6bt

noting at such a time t the  dy/dt = 0 by assumptions above.  
The second derivative of this equation is

18at−12a−18bt+6b

where at such time t we supply an additional parameter k equal to such equation.

Thus the second derivative equation at time t solved for t becomes eq.(1)

12a−6b+k18a−18b

                 
The first order equation at time t solved for b becomes eq.(2)

[a2−16−27a2+k2−−−−−−−−−√,a2+16−27a2+k2−−−−−−−−−√]

Assuming the the first equation in the equations set (2) for b implies with back substitution to eq.(1)

9a+k+−27a2+k2−−−−−−−−−√9a+3−27a2+k2−−−−−−−−−√

for time t providing eq.(3).

Thus we can see how large our original poly becomes with substitutions of eq.(3) and b from eq.(2)
into the parametric equation



1(3a+−27a2+k2−−−−−−−−−√)3(a9(−k+2−27a2+k2−−−−−−−−−√)2(9a+k+−27a2+k2−−−−−−−−−√)−p(3a+−27a2+k2−−−−−−−−−√)3+154(3a−−27a2+k2−−−−−−−−−√)(−k+2−27a2+k2−−−−−−−−−√)(9a+k+−27a2+k2−−−−−−−−−√)2)

here it appears a numeric method will be needed for roots solutions