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Would you please help me to prove or disprove that “If a control polygon is symmetric about the X-axis then also the Bézier curve defined by this control polygon is symmetric about the X-axis”, (Bézier curves of either odd or even degree can be considered). I know Symmetry property of Bézier curve but I don’t know how to fit X-axis symmetry into the Bézier Symmetry formula to prove the given question or is there any other way?

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Suppose we have a cubic Bézier curve with control points $\mathbf{P}_0 =(x_0,y_0)$, $\mathbf{P}_1 =(x_1,y_1)$, $\mathbf{P}_2 =(x_2,y_2)$, $\mathbf{P}_3 =(x_3,y_3)$. If the control points are symmetric with respect to the $x$-axis, this just means that $y_0 = -y_3$ and $y_1 = -y_2$.

Let $\mathbf{C}(t) = \bigl( x(t),y(t)\bigr)$ be the curve defined by these four control points. To show that this curve is symmetric with respect to the $x$-axis, we have to show that $y(t) = -y(1-t)$. But we have $$ y(t) = (1-t)^3y_0 + 3t(1-t)^2y_1 + 3t^2(1-t)y_2 + t^3y_3 $$ and $$ -y(1-t) = -t^3y_0 - 3t^2(1-t)2y_1 - 3t(1-t)^3y_2 - (1-t)^3y_3 $$ A little algebra, using the fact that $y_0 = -y_3$ and $y_1 = -y_2$ will show you that these two expressions are in fact the same, so the curve is symmetric with respect to the $x$-axis.

The same sort of technique will work with any curve of odd degree.

Curves of even degree can be treated similarly; the symmetry condition tells you that the middle control point must lie on the $x$-axis, and the other control points can be handled as in the odd degree case.

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