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?

  • $\begingroup$ Can we consider $a=0$? In this case the intersection will be a circle We can't just substitute one equation into another and make the first disappear. We need to convert the system of two equations into another system of also two equations $\endgroup$ Sep 17, 2021 at 8:59
  • $\begingroup$ The case where $a=0$ is some sort of the trivial solution, I'm looking for when the intersection is an ellipse with $e>0$. And, since every point on the intersection makes both equations true, it also makes the third one. Due to symmetry, the intersection can be seen 'from the side' as $z(x)$, which is the third one. $\endgroup$
    – omrir
    Sep 17, 2021 at 9:05
  • $\begingroup$ Cone and cylinder must also satisfy conditions II and III in that paper: I don't think having parallel axes is enough. $\endgroup$ Sep 17, 2021 at 9:22
  • $\begingroup$ In fact, $a=0$ would not result in a solution at all, would it? That intersection would be not one circle, but rather two circles in the parallel planes of $cz^2-b^2=0$. That is assuming $b\ne0$ and $c\gt0$. Without those conditions we would not have a cone and a cylinder. $\endgroup$
    – Pope
    Sep 17, 2021 at 13:09
  • $\begingroup$ The intersection you almost always get is a (non-planar) irreducible smooth genus one curve, looking like two ovals over the reals and like a torus over the complexes, reducible only when it's two (each but not as a union planar) circles for $a=0.$ $\endgroup$ Sep 18, 2021 at 7:25

4 Answers 4


In this answer, I derive a formula for the eccentricity of an ellipse in terms of angles made by the surface of the cone, and the cutting plane, with a "horizontal" plane (perpendicular to the axis). Complementing (for the sake of easier description here), we can restate the formula as $$e = \frac{\cos\theta}{\cos\phi} \tag1$$ where $\phi$ is the angle made by (a generator of) the cone and its axis, and $\theta$ is the (non-obtuse) angle made by the cutting plane (or, more-specifically, the conic's major axis) and the cone axis. It easy to show that, for the case of a cylinder, the corresponding relation matches the result of setting $\phi=0$ (and writing $\theta'$ to distinguish from $\theta$): $$e = \cos\theta' \tag2$$ So, if a cone and cylinder have a planar (thus, elliptical) intersection, then the ellipse's major axis makes angles $\theta$ and $\theta'$ with their respective axes, such that $$\cos\theta'=\frac{\cos\theta}{\cos\phi} \tag3$$ Thus, given any plane cutting a cone in an ellipse, we can find an appropriate cylinder that has the ellipse as its intersection. (Technically, the ellipse may only be part of the overall intersection.) Knowing $\phi$ and $\theta$, we use $(3)$ to find $\theta'$. (Note: Given that the plane cuts the cone in an ellipse, we know $1>e=\cos\theta'$ so that $\theta'$ is defined.) This angle determines the cylinder's axis through the ellipse's center; in fact, for $\theta'\neq \pi/2$, we have two possibilities for that axis. The desired cylinder's radius is necessarily the ellipse's minor radius. So, generally, there's are two families of configurations (essentially parameterized by $\theta$ and a scale factor, and the choice of cylinder axis) of cones and cylinders having (partially) planar intersection, with no requirement that the cone and cylinder have parallel axes.

That said, in the particular case where the cone and cylinder do have parallel axes, then $\theta=\theta'$. Assuming $\phi\neq 0$ (the cone is non-degenerate), then the only way to satisfy $(3)$ is with $\cos\theta=\cos\theta'=0=e$: the intersection is a circle. Clearly, the cylinder's axis necessarily coincides with the cone's. An "offset" configuration ($a\neq 0$ in the question) is invalid, re-confirming conclusions made in other answers.


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 $a=0$.

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The mistake in the paper occurs in the very first equation. In fact,

$$ b \neq \frac{r + R}{2}. $$

The minor axis of the ellipse lies in a plane halfway between the plane of the circle of radius $r$ and the plane of the circle of radius $R,$ and that plane intersects the cone in a circle of radius $(r + R)/2.$ But the minor axis of the ellipse is not a diameter of that circle. The center of the ellipse is at a distance $(R - r)/2$ from the center of the circle. Therefore

$$ b^2 = \left(\frac{r + R}{2}\right)^2 - \left(\frac{R - r}{2}\right)^2 = rR. $$

That is, $b = \sqrt{rR},$ so $b$ is the geometric mean of $r$ and $R$, not their arithmetic mean.

Moreover, from this answer we know that a given ellipse can be produced by a cone whose vertex lies anywhere on a certain hyperbola in a plane perpendicular to the plane of the ellipse, and that the axis of the cone must be tangent to the hyperbola. The angle that the axis makes with the plane of the ellipse therefore varies depending on which point on the hyperbola is the vertex. On the other hand, there are exactly two cylinders whose intersection with the plane is the given ellipse, and the angle the axis of one of these cylinders makes with the plane is determined by the ratio $a/b.$

So in general, given a cone that intersects a plane in an ellipse, the axis of the cylinder that intersects the same plane in the same ellipse will not be parallel to the axis of the cone. The cylinder is the limiting case of a cone produced by a point on the hyperbola very far from the plane. The angle between the cone's axis and the cylinder's axis can approach but never quite equal zero.


The intersection of two quadric surfaces (including cones and cylinders) is planar if and only they are tangent to a common third quadric. In the case of cylinders and cones, this third quadric will always be a sphere, I think. More here.


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