Show that the following system of parametric equations describes a line or a parabola: $$\begin{cases} x=a_1t^2+b_1t+c_1 \\ y=a_2t^2+b_2t+c_2 \end{cases}, t\in\mathbb{R}.$$

  • $\begingroup$ It may also describe a single point. $\endgroup$ – lhf Oct 18 '13 at 3:52
  • $\begingroup$ Or a straight line. $\endgroup$ – Michael Hoppe Oct 18 '13 at 7:13

You can eliminate the parameter $t$: Subtraction of the two equations gives: $$a_2 x-a_1 y=(a_2b_1-a_1b_2)t+(a_2c_1-a_1c_2)$$ Multiply first equation by $(a_2b_1-a_1b_2)^2$: $$(a_2b_1-a_1b_2)^2x=a_1 ((a_2b_1-a_1b_2)t)^2 +b_1 (a_2b_1-a_1b_2)^2 t+(a_2b_1-a_1b_2)^2c_1$$ from which: $$(a_2b_1-a_1b_2)^2x=a_1 (a_2 x-a_1 y-(a_2c_1-a_1c_2))^2 +b_1 (a_2b_1-a_1b_2)^2(a_2 x-a_1 y-(a_2c_1-a_1c_2))+(a_2b_1-a_1b_2)^2 c_1.$$ Rewrite in the form: $$ax^2+bxy+cy^2+ux+vy+w=0$$ then it's enoght to check $b^2-4ac=0$ to verify that this is a parabola (provided that $a,b,c$ not all $0$).

  • 2
    $\begingroup$ you missed some squares in your second line $\endgroup$ – user72870 Oct 17 '13 at 22:34
  • $\begingroup$ @user72870 Fixed missing squares, thank'you. $\endgroup$ – Fabio Lucchini Jan 2 '18 at 19:57

There's a standard approach to this type of thing (which goes way beyond simple quadratics) called the "method of resultants" to remove parameter $t$ from your system.

Computer algebra systems make the process straightforward. In Mathematica, for instance, one simply calls

Resultant[-x + a1 t^2 + b1 t + c1, -y + a2 t^2 + b2 t + c2, t]

to get

$$0 = a_2^2 x^2 - 2 a_1 a_2 x y + a_1^2 y^2 + \cdots = A x^2 + 2 B x y + C y^2 + \cdots \qquad (\star)$$

such that $B^2 - A C = 0$ implies that the solution set is a parabola (or some degenerate form depending on whether one or more of $A$, $B$, $C$ vanish).

The rationale for the elimination technique, as it was explained to me, is ingenious. One simply observes that a polynomial system in powers of a single unknown $t$ can be interpreted as a linear system in terms of multiple unknowns representing the powers of $t$.

In this problem, we start by introducing the unknowns $t_1$ and $t_2$ (which correspond to $t$ and $t^2$). The original system becomes

$$\begin{align} x &= a_1 t_2 + b_1 t_1 + c_1 \\ y &= a_2 t_2 + b_2 t_1 + c_2 \end{align}$$

(After all, for any $t$, the values $t_1 := t^1$ and $t_2 := t^2$ happen to be numbers that satisfy the system, right?)

Okay, so we have two equations in two unknowns. Let's solve:

$$t_1 = -\frac{a_2 x - a_1 y - c_1 a_2 + c_2 a_1}{a_1 b_2 - a_2 b_1}\qquad t_2 = \frac{b_2 x - b_2 y + b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$$

Now, recall that ---oh, yeah!--- $t_1 = t^1$ and $t_2 = t^2$. Since $(t^1)^2 = t^2$, it must be that $(t_1)^2 = t_2$, so that we can write: $$\left(-\frac{a_2 x - a_1 y - c_1 a_2 + c_2 a_1}{a_1 b_2 - a_2 b_1}\right)^2 = \frac{b_2 x - b_2 y + b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$$

Manipulate as needed, and we arrive at $(\star)$. No more $t$s ... Achievement unlocked!

Interestingly, the process works for polynomial systems of arbitrary (and not-necessarily-matching) degree. For instance, consider $$\begin{align} x &= a_4 t^4 + a_3 t^3 + a_2 t^2 + a_1 t + a_0 \\ y &= \phantom{a_4 t^4 + a_3 t^3 + \;} b_2 t^2 + b_1 t + b_0 \end{align}$$

We begin as before, introducing unknowns $t_1$, $t_2$, $t_3$, $t_4$ to get this linear system: $$\begin{align} x &= a_4 t_4 + a_3 t_3 + a_2 t_2 + a_1 t_1 + a_0 \\ y &= \phantom{a_4 t_4 + b_3 t_3 + \;\;} b_2 t_2 + b_1 t_1 + b_0 \end{align}$$

Now we solv---- oh, wait. We have four unknowns but only two equations. That's hardly helpful.

Well ... The trick for dealing with this (and accommodating issues such as the possibly-vanishing denominator $a_1 b_2 - a_2 b_1$ in the degree-2 system above) is also ingenious, but it's a bit more TeX work than I have time for at the moment. I'll have to return to this later. (Actually, since this goes way beyond the scope of this particular problem, I'll probably relegate the full description to a Bloog post and link to it.)

  • $\begingroup$ There's a lot more about resultants, elimination and implicitization in the Sederberg notes that I linked to in my answer. People around here often ask if algebraic geometry is any use for anything. This discussion shows that the old stuff is useful, at least. I can't say about the new stuff, because I don't understand any of it :-) $\endgroup$ – bubba Oct 18 '13 at 9:51
  • $\begingroup$ @bubba: Ah. I hadn't gotten far enough into the Sederberg notes to see the discussion of the linear equation approach (starting in section 17.4 of those notes, for anyone interested). The beginning stuff was rather opaque to me (as is Wikipedia's entry on the method of resultants), which is why I launched into my own exposition of the rationale. (Thanks to Sederberg ---and you--- I can save myself some effort! :) This old stuff is definitely useful; the method of resultants is the go-to equation-solving technique in my research, and I reference resultants in many of my M.SE answers. $\endgroup$ – Blue Oct 18 '13 at 18:28

Let $u=a_2 x-a_1 y$, which is of the form $at+b$.

Let $v=b_2 x-b_1 y$, which is of the form $ct^2+d$.

This means that in the coordinates $(u,v)$ the curve is a point, a line, or a parabola. For instance, when $a\ne 0$ and $c\ne0$, we have $v=c(u-b)^2/a^2+d$, which is parabola.

  • $\begingroup$ The first approach is very nice. Thanks. The second approach seems to be circular reasoning, to me. The given curve is certainly a quadratic Bezier curve, but how do we know that a quadratic Bezier curve is a parabola? By going through the reasoning shown in the various answers given here, I think. $\endgroup$ – bubba Oct 18 '13 at 4:58
  • $\begingroup$ @bubba. I think you're right. I've edited my answer. Thanks. $\endgroup$ – lhf Oct 18 '13 at 11:36

Hint: Prove that $x(t)$ and $y(t)$ satisfy together a quadratic equation of the form $$u_{00}+u_{10}x+u_{01}y+u_{20}x^2+u_{11}xy+u_{20}y^2=0$$ so that it will result in a quadratic curve.

Now look at the tangents at each $t$, we have $x'(t)=2a_1t+b_1$ and $y'(t)=2a_2t+b_2$. If $t\to\infty$, the tangent $\displaystyle\frac{y'(t)}{x'(t)}$ will tend to $\displaystyle\frac{a_2}{a_1}$. We get the same for $t\to -\infty$.

You also need to check the cases when $a_1=0$ or $a_2=0$ separately, but else this curve has exactly one direction in the infinity, so it must be a parabole (a hyperbole would have two different limit directions in the edge of the plane).


As the other two solutions indicate, the first step is to "implicitize" the given equations. You can do this from first principles, as in Fabio's answer, but there are also some established techniques based on "resultants". Resultants were popular in the 19th century, but they were forgotten until they were resurrected by CAD people in the 1980's. You can find a discussion of the relevant techniques in these notes by Tom Sederberg.

Once you have an implicit equation, you can just check that it's discriminant (the "$b^2 - 4ac$" thing) is zero.

This whole approach seems rather roundabout, to me. It ought be possible to identify a curve as a parabola without going to all the trouble of implicitizing, but I don't know how to do this.

  • $\begingroup$ See my answer... $\endgroup$ – lhf Oct 18 '13 at 3:56
  • $\begingroup$ "... forgotten until resurrected by CAD people in the 1980's." Interesting. I was introduced to resultants ---specifically, Sylvester's Eliminant--- in 1986 by a curmudgeonly professor on the very brink of retirement. Given his age, I wouldn't've pegged him as a computer guy (then again, I knew nothing of him outside the lecture hall). It's kinda cool to think that he could have been either part of this resurrection movement, or else a keeper of the flame for this otherwise-forgotten technique. (Looking back, the theme of that entire course seemed more flame-keepy than resurrecty.) $\endgroup$ – Blue Oct 18 '13 at 9:18
  • $\begingroup$ I'd guess he was a flame-keeper. The guys who brought it back were people like Tom Sederberg, Ron Goldman, and others. At that point, many people in the CAD community believed that it was impossible to find implicit equations for things like parametric cubic surfaces. You might find this interesting: mathoverflow.net/questions/143914/… $\endgroup$ – bubba Oct 18 '13 at 9:46
  • $\begingroup$ @bubba: Thanks for the pointer to the discussion about Weil. I wasn't aware of that history. $\endgroup$ – Blue Oct 18 '13 at 18:32

Caveat: I'm not sure that this is correct. But, if it's not, perhaps someone can salvage the argument.

Let $$ \mathbf{M} = \left[\matrix{ a_1 & b_1 \\ a_2 & b_2}\right] \quad ; \quad \mathbf{c} = \left[\matrix{ c_1 \\ c_2}\right] $$ Then the equations can be written $$ \mathbf{x} = \mathbf{M}\left[\matrix{ t^2 \\ t}\right] + \mathbf{c} $$ So, the curve is the image under the affine transformation $\mathbf{x} \mapsto \mathbf{M}\mathbf{x} + \mathbf{c}$ of the parametric curve $t \mapsto (t^2,t)$, which is obviously the parabola $y^2 = x$.

Now the shaky part -- affine maps preserve ratios of distances, and a parabola is defined by a ratio of distances, so affine maps preserve parabolas. Therefore the given curve is a parabola.

The trouble is that the definition of a parabola involves a distance to a line, and I'm not sure what happens to this under an affine transformation.

But, if this argument is actually correct, then it avoids all the implicitization tedium, which is nice.

  • $\begingroup$ Unfortunately, the parabola’s focus and directrix don’t get mapped to the focus and directrix of the parabola’s image. The image of the directrix isn’t even parallel to the new directrix in general. It looks like the axis gets mapped to a line that’s parallel to the new axis, though. I think this approach can work, though. Since an affine transformation is involved, perhaps moving to the projective plane might be the key. $\endgroup$ – amd Dec 12 '16 at 8:34
  • $\begingroup$ Examine the quadratic part of the matrix in the vector form of $y^2=x$ under an affine transformation. Its determinant will vanish, which means that the image is a parabola, perhaps degenerate. This isn’t entirely in the spirit of this answer, since the equation of the source parabola is made explicit, but it does use the idea of the parametric curve’s being a transformed standard parabola. I’ve added an answer along these lines. $\endgroup$ – amd Dec 12 '16 at 10:34

Using the idea in one of bubba’s answers, let $$\mathbf A=\begin{bmatrix}a_1&b_1&c_1 \\ a_2&b_2&c_2 \\ 0&0&1\end{bmatrix}$$ so that the parametric equation can be written as $$\mathbf x=\mathbf A\begin{bmatrix}t^2\\t\\1\end{bmatrix}.$$ The curve is therefore the image under an affine transformation of $(t^2,t)$, which in Cartesian form is the parabola $y^2=x$. The latter equation can be written in matrix form as $$\mathbf x^T\mathbf Q\mathbf x=\begin{bmatrix}x&y&1\end{bmatrix}\begin{bmatrix}0&0&-\frac12\\0&1&0\\-\frac12&0&0\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix}=0.$$ The determinant of the upper-left $2\times2$ submatrix is zero, which identifies this as a parabola. To apply this same transformation to $\mathbf Q$, we need to compute $\mathbf Q'=(\mathbf A^{-1})^T\mathbf Q\mathbf A^{-1}$. If $\det\mathbf A=a_1b_2-a_2b_1=0$, it’s not hard to show that the parametric equation describes a line or a point instead of a parabola. We’re really only interested in the determinant of the upper-left submatrix of $\mathbf Q'$: since the determinant of this block of $\mathbf Q$ is $0$, we know that it, too, vanishes. So, the image of $y^2=x$ under an affine transformation is also a parabola. Since we’re assuming that $\mathbf A$ is nonsingular, neither is $(\mathbf A^{-1})^T\mathbf Q\mathbf A^{-1}$, so this parabola is non-degenerate.

An “implicitized” equation for the parabola can be read directly from $\mathbf Q'$. Note that since $(\mathbf A^{-1})^T\mathbf Q\mathbf A^{-1}=(\det\mathbf A)^{-2}\operatorname{adj}(\mathbf A)^T\mathbf Q\operatorname{adj}(\mathbf A)$ we can factor out the determinant of $\mathbf A$ by performing this transformation with its adjugate instead of its inverse. This will make the calculations a bit simpler.


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