Parabola in parametric form

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 $0$).

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.)

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.

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).