It seems to me that a parabola on any plane (not just the standard 2D xy-plane) should follow the standard properties of a parabola. By definition, a parabola is the set of points such that any point, $P$, on the parabola, is equidistant from the focus, $F$, and the directrix.
Consider the parabola formed as a result of intersecting the plane
$-y + z -5 = 0$ and the 1-sheeted hyperboloid $\frac{x^2}{2^2}+\frac{y^2}{1^2}-\frac{z^2}{1^2} = 1$.
The conic produced in general $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ form is simply $0.025x^2 + (-1)y + (-2.6) = 0$. Since $B^2-4AC= 0^2-4A*0 = 0$, we see that this is a parabola. And a parabola in general form is $f(x) = ax^2 + bx + c$ where in this case $a=0.025, b=0, c=-2.6$.
The vertex, $V$, is $(\frac{-b}{2a}, \frac{4ac-b^2}{4a}) = (0,-2.6)$.
The focus, $F$, is $(\frac{-b}{2a},\frac{4ac-b^2+1}{4a}) = (0,7.4)$.
The directrix $y = \frac{4ac-b^2-1}{4a}$ is $y=-12.6$.
Now, if I project these points onto the intersecting plane, to view the 3D curve of intersection I get $V=(0,-2.6,2.4)$ and $F=(0,7.4,12.4)$.
And the directrix is the line $y=-12.6,z=-7.6$ which I can express as $(0,-12.6,-7.6)+(k,0,0)$ $\forall$ $k$ $\in \mathbb{R}$ where $\mathbb{R}$ is the reals. You may verify the line and these points lie on the plane.
Now let's check the 3D distance from a point on the 3D curve of intersection to the 3D directrix and 3D focus. You may verify that $P=(10,-0.1,4.9)$ satisfies the equation of the hyperboloid and the plane, so let's check that it upholds the properties of a parabola given above.
The distance to $F$ is $\sqrt{(10-0)^2+(-0.1-7.4)^2+(4.9-12.4)^2} \approx 14.577$.
The orthogonal distance to the directrix is $\sqrt{(10-10)^2+(-0.1-12.6)^2+(4.9-7.6)^2} \approx 17.678$.
Since $14.577 \neq 17.678$, clearly I am either misapplying basic concepts or do not understand the basics of a parabola.
Could someone please explain my misunderstanding or my error? Thank you.