Parametric 3D Parabola Directrix

The midpoints of the points produced by $t$ and $-t$ of $p(t) = p_0 + p_1t + p_2t^2$ are $m(t) = \frac{1}{2}(p_0 + p_1t + p_2t^2 + p_0 + p_1(-t) + p_2(-t)^2) = p_0 + p_2t^2$ , which is a straight line. As this is a unique property of the axis symmetry, the axis of symmetry must follow vector $p_2$ , which in this case is $(3, -1, 2)$ .

Since $p_0 = (4, 2, 3)$ , $p_1 = (1, -1, 4)$ , and $p_2 = (3, -1, 2)$ , and $p(t) = p_0 + p_1t + p_2t^2$ , the parabola has parametric equations $x = 4 + t + 3t^2$ , $y = 2 - t - t^2$ , and $z = 3 + 4t + 2t^2$ . These points will be on some plane in the form of $x + by + cz = d$ with no $t$ terms, so that $1 - b + 4c = 0$ (from the $t$ terms) and $3 - b + 2c = 0$ (from the $t^2$ terms). These two equations solve to $b = 5$ and $c = 1$ , so that the equation of the plane is $x + 5y + z = d$ , with a normal vector of $(1, 5, 1)$ .

The unit vector of the directrix $(p, q, r)$ must be perpendicular to $(3, -1, 2)$ so that $3p - q + 2r = 0$ , be perpendicular to $(1, 5, 1)$ so that $p + 5q + r = 0$ , and be a unit vector so that $p^2 + q^2 + r^2 = 1$ . These three equations solve to $(p, q, r) = (-\frac{11\sqrt{42}}{126}, -\frac{\sqrt{42}}{126}, \frac{8\sqrt{42}}{63}) \approx \boxed{(-0.5658,-0.0514, 0.8230)}$ for $r > 0$ .