I'm solving this problem about Cubic Bezier Curve which have $4$ control points $C_0,C_1,C_2,C_3$. $$ C_0 = (0,0),\: C_1 = (x_1,y_1),\: C_2=(x_2,y_2),\: C_3=(1,0) $$ where $0 < x_1 < x_2 < 1$ and $y_1, y_2>0$.
If $C_1$ and $C_2$ are symmetric with respect to axis $x = 0.5$ (such as $C_1 = (0.25,1)$, $C_2 = (0.75,1)$), obviously whole bezier curve is symmetric (w.r.t. $x=0.5$). But I'm still confusing if converse of this statement is still correct. Is any symmetric (cubic) bezier curve can be existed with two unsymmetrical control points $C_1,C_2$?
I tried with reduction, if $C_1, C_2$ is not symmetric and the image $B(C_0,C_1,C_2,C_3)$ is symmetric, there should be $C_1^\prime$ that symmetric with respect to $C_1$ but $C_1^\prime := C2 \cdot B(C_0,C_1,C_1^\prime,C_3)$ is symmetric and when I move just one control point $C_1^\prime$ to $C2$, I'm not sure that symmetric property will broke or not.