When is the intersection of a cone and a cylinder is planar? I'm trying to understand better the link between the algebraic and geometric interpretation of the ellipse, and I've wondered about the intersection of a right circular cylinder and a right circular cone.
On one hand, we know that the intersection of each with a plane (given that the plane is not too slanted) is an ellipse, and so we can tweak the planes and the radius of the cylinder to give the same ellipses, as can be seen here.
In this, they conclude that if the cone and cylinder's axes are parallel, and the cone axis fall within the cylinder, the intersection is a planar ellipse.
However, it does not work when dealing with the algebraic equations of the two:
$$\begin{cases}
(x-a)^2 + y^2 = b^2\\
x^2 + y^2 = cz^2
\end{cases}$$
from substituting the second into the first, we get $2ax=cz^2+a^2-b^2$, which is the equation of a parbolic cylinder. (The case for $a=0$ gives a circle, and it sort of the trivial solution).
Which approach is the correct one, and what's wrong with the other?

As you correctly found, the intersection of a cone and a cylinder with parallel axes lies on a parabolic cylinder, whose axis is perpendicular to the axes of cone and cylinder. This rules out a planar intersection, save for the trivial case $a=0$.

