The curve
(0,0) parabola (1,1)
is equivalent to
(0,0) .. controls (.5,0) and (0.8875,0.775) .. (1,1)
but the exact parabola is
(0,0) .. controls (1/3,0) and (2/3,1/3) .. (1,1)
We can see in the code of pgfcorepathconstruct.code.tex at line 1297:
{\pgfqpoint{.5\pgf@xc}{0\pgf@yc}}% found by trial and error
{\pgfqpoint{.8875\pgf@xc}{.775\pgf@yc}}% found by trial and error
By the way there is no reason with bend option to draw a parabola using two cubic curves : any parabola can be drawn by a single quadratic (and so cubic) curve. You can check my answer at TeX.SX (tex.stackexchange.com).
I think that bend=<point> should be interpreted as "the bend is over the x coordinate of the point". This is backward compatible when there exist a parabola that has a bend at the point. And in all other cases the current parabola operation is drawing a non parabola curve, so there is no need of backward compatibility.
Also, it will be good to add trough=<point> option.
Many thanks for the hint. I can confirm this and also added the correct values for the "first" parabola point in the definition (see attached code).
For the record: Finding better solutions also for the
sinandcosfunctions with thebezierplotpackage was not successful as can also be seen in the attached code.For your suggestion to the improvement of the
parabolafunctionality: I created feature request #117.If you want a "better" approximation for
sinyou can keep0.31831which is1/pibut you can replace0.63503by0.63662which is2/pi. And the corresponding values forcosare0.36338which is1-2/piand0.68169which is1-1/pi. But it is not clear that this is a "better" approximation ... it depends how you define "better".Regarding the
sin/cosimprovement see patch request #26.