Parabola in parametric form Show that the following system of parametric equations describes a line or a parabola:
$$\begin{cases} x=a_1t^2+b_1t+c_1 \\ y=a_2t^2+b_2t+c_2 \end{cases}, t\in\mathbb{R}.$$
 A: 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 $0$).
A: 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.
A: 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.)
A: 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).
A: 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.
A: 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.
A: 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.
