How to prove axis of symmetry for Bezier-Bernstein curve?

Question

I have a planar bezier curve of degree $\ n=3 $ with the control points:

$\ p_{o}=[0, 0]^T $

$\ p_{1}=[1, 1]^T $

$\ p_{2}=[2, 1]^T $

$\ p_{3}=[3, 0]^T $

Suppose I have the bernstein polynomials. How can I prove that the curve has axis of symmetry the line $\ x=\frac{3}{2} $ ? I thought I should find the value $\ t_{o} $ for which I get $\ P(t_{o})=\frac{3}{2} $, using the $\ x $ coords of the control points but after that I have no clue.

Answer 1

Plugging your control points (x, y) values to the cubic Bezier formula, you will get

$x(t)=3t$
$y(t)=3t-3t^2$

Therefore, we can rewrite this curve in explicit form as

$y(x)=x-x^2/3$

To prove that the $y(x)$ is symmetric to $x=3/2$, we need to prove that $y(3/2-x)=y(3/2+x)$. This can be easily done after some simple algebra, which I will leave it for your own exercise.

Answer 2

Although it's more long-winded, I think there's also more insight from a general approach. Both of these lemmata can be proven algebraically from the Bernstein polynomials or geometrically with de Casteljau subdivision. In my opinion the geometric approach is more instructive.

Let $B(t, P_0, \ldots, P_n)$ be the interpolated point at parameter value $t$.

Lemma 1: if $\alpha$ is an affine transform then $$B(t, \alpha P_0, \ldots, \alpha P_n) = \alpha B(t, P_0, \ldots, P_n)$$ The key to proving this is that the basis functions always sum to unity.

Lemma 2: reversibility $$B(t, P_0, \ldots, P_n) = B(1-t, P_n, \ldots, P_0)$$

With both lemmata in hand, you can take the affine involution $\alpha$ which reverses the control points of your curve and say $$B(t, p_0, \ldots, p_n) = B(1-t, p_n, \ldots, p_0) \\ = \alpha^{-1} B(1-t, \alpha p_n, \ldots, \alpha p_0) \\ = \alpha B(1-t, p_0, \ldots, p_n)$$ i.e. if the control points are symmetric, the whole curve is symmetric.