Given a general quadratic, where not all of the coefficients are 0:
$$ \begin{array}{lr} Ax^2 + By^2 + Dxy + Gx + Hy + J = 0 & (1) \end{array}$$
We can think of this as a map from the set of viable coefficients in ${\rm I\!R}^6$ to the set of manifolds in ${\rm I\!R}^2$.
Restricting our equation to a parabola restricts the angle at which the plane can intersect the cones, and so this results in a variable that is not "free". Formally, this is relatively easy to obtain as $D^2 - 4AB = 0$. So, we could set $A = D^2/(4B)$ and "do away" with one of the variables, obtaining a map instead from ${\rm I\!R}^5$ to the set of manifolds on ${\rm I\!R}^2$.
In 3D, the equivalent thing is a quadric surface, the points of which satisfy the following general equation:
$$ \begin{array}{lr} Ax^2 + By^2 + Cz^2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J = 0 & (2) \end{array}$$
I suspect that constraining this equation to be a paraboloid gives a similar restriction on the variables, but it is not entirely clear how I might perform a substitution to "do away" with any of the variables. Is there a way that I can rewrite equation (2) in only 9 coefficients to constrain the problem to paraboloids?