I'm trying to find the value(s) of the parameter $t$ at the intersection point(s) between a 2D general parabola (as a parametric function of $t$) and a line whose equations can be derived from two points $p$ and $q$ ($p \not = q$). The parabola is defined below:
$$x(t) = \frac{1}{2} a_x t^2 + v_x t + x_0 \\ y(t) = \frac{1}{2} a_y t^2 + v_y t + y_0$$
My Interest:
If I could find this solution, then I could predict the time-to-collision between two non-rotating, rigid line segments. By using the relative motion between them as the $a$ and $v$ in this parabolic curve for each point, and finding the intersection between it and the other line segment, I can find the smallest $t > 0$ such that the lines are colliding.
This would be for a physics engine, and using acceleration in the calculation would allow for some huge efficiency gains (not having to constantly predict and handle collision between every pair of line segments, only during relevant time frames or when acceleration changes).
Edit:
Thanks to Andrea Mori, I was able to solve for this equation (given the line defined by $p$ and $q$):
$$(q_y - p_y) x(t) - (q_x - p_x) y(t) - |p \times q| = 0$$
(By treating $a_x b_y - a_y b_x$ as equivalent to $\left| a \times b \right|$ for 2D). Then, substituting $x(t)$ and $y(t)$, I was able to simplify down to more cross products:
$$\frac{1}{2}\left(\left| a \times q \right| - \left| a \times p \right|\right) t^2 + (\left| v \times q \right| - \left| v \times p \right|)t + (\left| r \times q \right| - \left| r \times p \right| - \left| p \times q \right|) = 0$$
The quadratic formula can then be applied using the clearly visible $a$, $b$, and $c$ values (though writing it out would get ugly). A negative discriminant ($b^2 - 4ac$) indicates no collision, zero indicates exactly one, and positive indicates two (the smallest $t > 0$ being the most important one).
I think the second formula would then translate to 3D, though I don't know without evaluating it some more.