On the Equation of the Ellipse Formed by Intersecting a Plane and Cone

Question

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.

Answer 1

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.

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 $\theta$: $$ \alpha={\pi+\gamma-\beta\over2}-\theta. $$

Answer 2

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

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).

• Why slicing a cone gives an ellipse [3Blue1Brown]