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This is a two-part question that arose while investigating properties of parabolas.

A parabola can be parameterized as $P:t\mapsto(X(t),Y(t))$, where $$\begin{align}X(t)&=at^2+bt+c \\ Y(t)&=At^2+Bt+C\end{align}$$ with $a$ and $A$ not both zero. This parameterization is not unique. For any two points $P_0$ and $P_2$ on the parabola, the Bernstein polynomial $$(1-t)^2P_0+2(1-t)tP_1+t^2P_2,\tag{*}$$ where $P_1$ is the intersection of the tangents at $P_0$ and $P_2$, is a distinct parameterization of the parabola. (For $0\le t\le 1$, this is the quadratic Bézier curve with control points $P_0$, $P_1$ and $P_2$.) What I was wondering is if there are any others. That is, are there any parameterizations of this form for a non-degenerate parabola that can’t be represented as the Bernstein polynomial (*) generated from a pair of points on the curve?

Secondly, it’s evident that any such parameterization be converted into another via a transformation of the parameter, but what does this transformation look like? For the Bernstein polynomials, the parameter can be interpreted as the fractional displacement along line segments, so I suspect that they are related by a affine transformations of the parameter.

(I have my own solution to this, which I’ll post later, but I’d be interested in seeing other peoples’ takes on these questions.)

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  • $\begingroup$ A key idea for treating these questions is to use the fact that if $$N:=\pmatrix{0 & 1 & 0\\0& 0 & 1\\0&0&0}$$ then $$exp(tN)=\pmatrix{1 & t & t^2/2\\0&1&t\\0&0&1}$$ $\endgroup$ – Jean Marie Dec 11 '16 at 7:40
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The key point, I think, is that $\{1,t,t^2\}$ and $\{(1-t)^2,2t(1-t),t^2\}$ are both bases for the vector space of polynomials of degree two. Therefore, any polynomial of degree two has a unique representation as a linear combination of $\{1,t,t^2\}$, and a unique representation as a linear combination of $\{(1-t)^2,2t(1-t),t^2\}$. The second form is known as a "Bezier curve".

Changing basis doesn't change the curve, of course; it only changes its coefficients.

There are other interesting bases, too: Lagrange and Chebyshev are two examples.

If you use a linear fractional transformation to reparameterize, you can get a rational quadratic representation of a parabola. This won't change the shape of the curve, but will change the "speed" at which it is traversed.

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  • $\begingroup$ I knew there was something simple going on. I was too focused on the geometry to see it. $\endgroup$ – amd Dec 15 '16 at 18:15
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Observe first that this parameterization is regular: if $P'$ vanishes for some $t$, then the parabola is degenerate, i.e., $P$ represents the straight line $A(x-c)-a(y-C)=0$ (or even a single point!).

For a second-order Bézier curve $B_{P_0P_1P_2}(t)$, the outer control points, which correspond to $t=0$ and $t=1$, are on the parabola, while the middle control point is the intersection of the tangents to the parabola at these two points. We have $P(0)=(c,C)$ and $P(1)=(a+b+c,A+B+C)$. The equation of the tangent to the parabola at $t$ is $$(Y'(t),-X'(t))\cdot(x,y)=(Y'(t),-X'(t))\cdot P(t).$$ For our two points, this generates the system $$\begin{align}Bx-by&=Bc-bC \\ (2A+B)x-(2a+b)y&=(2A+B)(a+b+c)-(2a+b)(A+B+C)\end{align}$$ with solution $$x=\frac b2+c, y=\frac B2+C.$$ The Bernstein polynomial generated by these control points is $$\begin{align}x&=(1-t)^2c+2(1-t)t\left(\frac b2+c\right)+t^2(a+b+c)\\ &=\left(a+b+c-2\left(\frac b2+c\right)+c\right)t^2+\left(2\left(\frac b2+c\right)-2c\right)t+c \\ &=at^2+bt+c\end{align}$$ and similarly for $y$. Thus, second-order Bézier curves account for all quadratic parameterizations of a parabola.

For the second part of the question, we can, without loss of generality, consider the parabola $x=\alpha y^2$ $(\alpha >0)$, with the parameterization $P:t\mapsto(\alpha t^2,t)$. The tangent to this parabola at $P(t)$ is given by the equation $x-2\alpha ty+\alpha t^2=0$, so the tangents at $P(t_1)$ and $P(t_2)$ intersect at the point $\left(\alpha t_1t_2,\frac{t_1+t_2}2\right)$. The corresponding Bernstein polynomial is $$(1-t)^2(\alpha t_1^2,t_1)+2(1-t)t\left(\alpha t_1t_2,\frac{t_1+t_2}2\right)+t^2(\alpha t_2^2,t_2)$$ which simplifies to $$\begin{align}x&=\alpha (t_1(1-t)+t_2t)^2\\y&=t_1(1-t)+t_2t.\end{align}$$ Observe that the coefficient of $t$ (the scale factor) is the directed length of the projection of the segment $\overline{P(t_1)P(t_2)}$ onto the $y$-axis. We thus conclude that for a fixed parabola, any quadratic parameterization of it can be converted to any other by an affine transformation of the parameter.

It’s also possible to prove this directly from the general parametric equations. The direction of the parabola’s axis is given by the vector $(a,A)$. An equation of the line through $P(t)$ parallel to this axis is $(-A,a)\cdot(x,y)=(-A,a)\cdot P(t)$ and $${(-A,a)\cdot P(t)\over\sqrt{a^2+A^2}}={(aB-Ab)t+(aC-Ac)\over\sqrt{a^2+A^2}}$$ is the directed distance of this line from the origin. If we invert this and subtract the distance of the axis from the origin, we end up with a normalized parameter that represents the signed distance of a point on the parabola from its axis. This can be interpreted as the $y$-coordinate of a frame in which the parabola’s axis is horizontal and its vertex is at the origin. The expression for the transformation between these parameterizations isn’t nearly as simple as in the previous paragraph, though.

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