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2D View of cone projection (blue line)

As shown in the figure above, I want to determine the equation of the ellipse formed by intersection of a tilted right cone and a plane. We know $\alpha,$ (angle formed with the vertical), $R$ as well as $\Omega$ (cone's half-angle). The author here Previously given solution proposed a solution which confuses me on the $x=(p_1-p_2)/2,$ $y=R\tan(\Omega)$ part...how do we know this point lies on the ellipse? The author did not justify this step clearly.

Richard
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  • It is quite obvious that the whole blue line is the projection of the cross section of the plane with the cone. Of course the linked solution does not prove that the cross section is an ellipse, it assumes this. – user Mar 27 '21 at 20:40
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    I chose the horizontal line as $x$-axis and in particular the origin as the midpoint of the blue segment (which is the major axis of the ellipse). $y$-axis is also horizontal, perpendicular to the page. If you draw, from the center of the cone base, a line parallel to $y$-axis, then it intersects the cone (and the ellipse) at points $\big((p_1-p_2)/2,\pm R\tan\Omega\big)$. – Intelligenti pauca Mar 27 '21 at 21:03
  • @Intelligenti pauca Thanks for the clarification. Another variation of this problem is the case of a rotated plane. John Alexiou suggests an intriguing setup in the answer below. – Richard Mar 27 '21 at 21:28
  • What do you mean by "rotated plane"? The answer below is using Dandelin's spheres, but it all depends on what you exactly want. – Intelligenti pauca Mar 27 '21 at 21:33
  • @Intelligenti pauca Ok. What I mean is that your solution would work in a 3D setup ( with xyz axes) where the the plane is $z=0.$ If the plane is now rotated about one of the axes, the ellipse formed is a 3D curve. This is actually what I'm working towards – Richard Mar 27 '21 at 21:40
  • I didn't use coordinates in my answer (except for the standard equation of the ellipse, just to save some time), hence I don't see how "rotating" the plane would change anything. You should explain more clearly what you are looking for, if you want to be helped. – Intelligenti pauca Mar 27 '21 at 21:48
  • Here's another answer of mine to a similar question: https://math.stackexchange.com/a/2351900/255730 – Intelligenti pauca Mar 27 '21 at 21:49
  • @Intelligentipauca I mean if the $x$-$y$ plane were to be rotated about any of the $x$,$y$ or $z$ axes while the vertex of the cone remains unchanged, the shape of the ellipse will certainly change. I want to represent the equation of the ellipse in parametric or cartesian equation if possible. I don't know if I'm making myself clear – Richard Mar 28 '21 at 00:46
  • I set up a 2D cartesian frame on the plane, just to write there the equation of the ellipse. But that has nothing to do with a possible 3D cartesian frame, which I never used. – Intelligenti pauca Mar 28 '21 at 08:43
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    If what you want is the equation of the ellipse in a 3D frame, then you must know it doesn't exist. You can describe an ellipse in 3D either as the intersection of two surfaces of which you give the equations (e.g. plane and cone), or as three parametric equations $(x(t),y(t),z(t))$. – Intelligenti pauca Mar 28 '21 at 08:55

2 Answers2

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Here's a 3-D picture, to explain more clearly that answer.

I chose as $x$-axis the intersection between the given plane, and a plane through the axis of the cone, perpendicular to it. I placed the origin at the midpoint of segment $AB$ (which is the major axis of the ellipse). $y$-axis is on the given plane, perpendicular $x$-axis. If you draw, from the center $M$ of the cone base, a line parallel to $y$-axis, then it intersects the cone (and the ellipse) at points $I$ and $H$, with coordinates $|x|=CM=(_1−_2)/2$, $|y|=MH=\tan Ω=\,$radius of the base of the cone.

enter image description here

EDIT.

To choose the intersecting plane such that the center of the ellipse is at a given point $C$ inside the cone, let's consider line $VC$, where $V$ is the vertex of the cone (see figure below).

If $C$ lies on the axis of the cone, then just choose a plane through $C$ perpendicular to the axis, to obtain as intersection a circle of centre $C$.

If $C$ doesn't lie on the axis, then consider the plane $\sigma$ through $C$ and the axis, intersecting the cone at two generatrices $VA$ and $VB$, and set $\angle CVB=\beta$, $\angle CVA=\gamma$, with $\beta>\gamma$. If a plane through $C$, perpendicular to $\sigma$, intersects $VA$, $VB$ at $A$ an $B$ respectively, then $AB$ is the major axis of the intersection ellipse, and $C$ is its center if $AC=BC$.

Let $\theta=\angle VCB$. From the sine law applied to triangles $VCB$ and $VCA$ one finds

$$ {BC\over VC}={AC\over VC}={\sin\beta\over\sin(\theta+\beta)}= {\sin\gamma\over\sin(\theta-\gamma)}, \quad\text{whence:}\quad \tan\theta={2\sin\beta\sin\gamma\over\sin(\beta-\gamma)}. $$ Finally, we can express angle $\alpha$ between the plane of the ellipse and a plane perpendicular to the axis as a function of $\theta$: $$ \alpha={\pi+\gamma-\beta\over2}-\theta. $$

enter image description here

Intelligenti pauca
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Not a solution

I remember seeing a YT video recently on this and it involved two spheres that are tangent to the inside of the cone and also the plane

fig1

Where these spheres touch the plane are the ellipse foci. From the two foci the equation of the ellipse can be found (or rather the parameters of the ellipse).

John Alexiou
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