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Throughout this post I'm assuming that the cone is pointing "up" in the $Z$ axis.

It's known that the intersection of a plane and cone is an ellipse. But what about the projection of this ellipse onto the $X,Y$ plane - is it always a circle? Or equivalently, does every intersection of a plane and cone lie in some cylinder pointing in the $Z$ direction? It is true for a plane and paraloboid intersection, but it doesn't seem like it for a cone plane intersection:

For a cone $z^2 = x^2 + y^2$ and a plane $z=\frac{x+1}{2}$ we have

$4 x^2 + 4 y^2 = 4z^2 = x^2 + 2x + 1$

$3x^2 -2x =-4y^2 + 1$

$x^2 - \frac{2}{3}x = \frac{-4}{3}y^2 + \frac{1}{3}$

$(x- \frac{1}{3})^2 - \frac{1}{9} = \frac{-4}{3}y^2 + \frac{1}{3}$

$(x- \frac{1}{3})^2 + \frac{4}{3}y^2 = \frac{4}{9}$

Which is not the equation for a circle.

Is there perhaps a cleaner way of showing that the projection of a plane and cone is not always a circle?

brainjam
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Jake1234
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    Your calculation looks algebraically elegant: Up to rotation, every non-vertical plane has equation $z = Ax + C$. If $A \neq 0$, equating the $z^2$s as you've done gives a diagonalized quadratic with $x^2$ and $y^2$ having respective coefficients $1 - A^{2}$ and $1$. Since these are unequal, the intersection does not project to a circle. – Andrew D. Hwang Jan 22 '22 at 21:27
  • I don’t think it’s true for paraboloid intersections, either. – bubba Jan 23 '22 at 00:41
  • @bubba At least for the basic $z=x^2 + y^2$ paraboloid I think it is true - this is known as the "lifting lemma" that is used in various Delaunay triangulation texts. – Jake1234 Jan 23 '22 at 16:10
  • @Jake1234. Interesting. Thanks. It’s hard to believe, and it seems to contradict the theorem I cited below (planar intersection iff common tangent quadric). In the case of paraboloid and cylinder, I wonder what this common tangent quadric is?? – bubba Jan 23 '22 at 23:44
  • @Jake1234. I read this. https://sites.cs.ucsb.edu/~suri/cs235/Lifting.pdf. Just high-school algebra, so hard to deny. But very surprising, geometrically. Thanks. – bubba Jan 23 '22 at 23:55

2 Answers2

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Suppose we have an elliptical intersection curve $E_1$ whose projection is a circle. Now we change the slicing plane a little bit — we rotate it around the minor axis of $E_1$, making its angle steeper. This will give us a new elliptical intersection $E_2$ that’s more elongated than $E_1$. $E_2$ will have the same minor radius as $E_1$, but a longer major radius.

When we project $E_1$ and $E_2$ onto a horizontal plane, we again get two ellipses that have the same minor radius but different major radii. So, they can’t both be circles.

Edit:
The statement in italics above is incorrect. Rotating the slicing plane around the minor axis changes the minor radius of the ellipse, as the following picture clearly shows.

enter image description here

bubba
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  • I don't think rotating the slice plane around the minor axis preserves the minor axis. – Jake1234 Jan 23 '22 at 15:59
  • @Jake1234. I’ll try it in a CAD system to check (unless you already did something similar). – bubba Jan 23 '22 at 23:48
  • The reasoning I had is that for a cone $z^2 = x^2 + y^2$ and the plane $z=1$, looking from the side (projection on $x,z$ axis), we have the center of the cut plane ellipse at a distance of $d=\sqrt{2}/2$ from the cone. Rotating the plane around the y axis, the center of the intersection ellipse comes closer to the cone surface, but there's a lower bound of around ~$d/2$. The center keeps going higher however, but stays at least $d/2$ away from the cone, so if we then look at the horizontal cut plane that contains the center of a steeper plane cone intersection, the minor axis must increase. – Jake1234 Jan 24 '22 at 03:53
  • @jake1234. You're right. My intuition failed me. It's pretty obvious when you look at a 3D model. – bubba Jan 24 '22 at 08:39
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If an intersection curve’s projection is a circle, then, as you pointed out, this curve must lie on a vertical cylinder that passes through the circle. So, the intersection curve must be the intersection curve $C$ of the cone and this cylinder.

If $C$ is a planar section of the cone, then it must be planar, obviously. It’s highly unlikely that intersecting a vertical cone and a vertical cylinder will give a planar curve. You get a planar curve if the cone and cylinder have the same centerline, but rarely (if ever) otherwise.

In fact, the intersection of two quadric surfaces is planar if and only if they are both tangent to some third quadric. See this question.

Also, see this question.

bubba
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