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There is a known ellipse

$ (r - r_0)^T Q_e (r - r_0) = 1 $

that is the intersection of an unknown right circular cone with the $xy$ plane.

Determine the equation of the cone, i.e. determine the position of its apex $A$, the direction of its axis, and its semi-vertical angle $\theta_c$.

My attempt:

There are six parameters here. Three for the position of the apex $A$, two for the axis direction, and one for the semi-vertical angle. An ellipse is determined by five parameters. So it would seem that there is one degree of freedom here. I could proceed by fixing one of the parameters of the cone, and solving for the other five. For example, I could assume that the elevation $z$ of the apex is given.

I am open to suggestions, hints, comments, and solutions.

Update: Based on a comment by Chris Lewis below, it would simply things if the ellipse is translated (shifted) and rotated so that it is centered at the origin of a coordinate system with its axes along the axes of the new coordinate system. To do this, diagonalize $Q_e$ into $Q_e = R_e D_e R_e^T $

Then we would have

$ (r - r_0)^T R_e D_e R_e^T (r - r_0) = 1 $

Now we introduce a change of variable by defining

$ u = R_e^T (r - r_0 ) $

With this the equation of the ellipse in the coordinate $u$ becomes

$ u^T D_e u = 1 $

1 Answers1

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As mentioned in the update in the question, I'll assume that the given ellipse has the form

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \tag{1}$

where $a \gt b$.

We know that the cone with such a section must have its apex in the $xz$ plane, so let the apex be at $A = (x_1, 0, z_1)$. Also, we know that the rotation matrix associated with the cone must a rotation matrix about the $y$-axis. Let the angle of rotation be $\theta_1$. Then

$ R = \begin{bmatrix} \cos \theta_1 && 0 && \sin \theta_1 \\ 0 && 1 && 0 \\ -\sin \theta_1 && 0 && \cos \theta_1 \end{bmatrix}\tag{2}$

In addition, the unrotated upright cone matrix takes the form

$ D_c = \begin{bmatrix} \cos^2 \phi && 0 && 0 \\ 0 && \cos^2 \phi && 0 \\ 0 && 0 && \cos^2 \phi - 1 \end{bmatrix} \tag{3} $

And the equation of the cone is

$ (r - A)^T Q_c (r - A) = 0 \tag{4}$

where $ r = [x,y,z]^T $, and $Q_c = R D_c R^T $

Calculating $Q_c = R D_c R^T $, we get,

$Q_c = \begin{bmatrix} \cos^2 \phi - \sin^2 \theta_1 && 0 && - \sin \theta_1 \cos \theta_1 \\ 0 && \cos^2 \phi && 0 \\ - \sin \theta_1 \cos \theta_1 && 0 && \cos^2 \phi - \cos^2 \theta_1 \end{bmatrix} \tag{5}$

Setting $r = [x, y, 0]$, will give the intersection of the cone with the $xy$ plane. For brevity, let $C = \cos \phi, s_1 = \sin \theta_1, c_1 = \cos \theta_1$. Then the equation of the intersection is

$( C^2 - s_1^2 ) (x - x_1)^2 + C^2 y^2 + (C^2 - c_1^2 ) (z_1)^2 - 2 s_1 c_1 (x - x_1)(-z_1) = 0 \tag{6}$

And this is a scalar multiple of the above given equation of the ellipse.

Equating coefficients,

$ - 2 x_1 (C^2 - s_1^2) + 2 z_1 s_1 c_1 = 0 \tag{7} $

$ C^2 - s_1^2 = \dfrac{k}{a^2} \tag{8} $

$C^2 = \dfrac{k}{b^2} \tag{9} $

$x_1^2 (C^2 - s_1^2) + z_1^2 (C^2 - c_1^2) - 2 z_1 x_1 s_1 c_1 = -k \tag{10} $

Substitute for $k$ from $(8)$ into $(9)$ and $(10)$, thus eliminating $k$.

$ - x_1 (C^2 - s_1^2) +z_1 s_1 c_1 = 0 \tag{11} $

$ C^2 b^2 = a^2 (C^2 - s_1^2) \tag{12}$

$ x_1^2 (C^2 - s_1^2) + z_1^2 (C^2 - c_1^2) - 2 z_1 x_1 s_1 c_1 = - a^2 (C^2 - s_1^2) \tag{13}$

with the aid of $(11)$ and $(12)$, equation $(13)$ becomes

$ - x_1^2 (C^2 - s_1^2) + z_1^2 (C^2 - c_1^2) = -a^2 (C^2 - s_1^2) = -b^2 C^2 $

hence,

$ \dfrac{x_1^2}{a^2} - \dfrac{ z_1^2 (C^2 - c_1^2) }{ b^2 C^2 } = 1 \tag{14}$

Now, note that $(C^2 - s_1^2) / C^2 = b^2 / a^2 $. Therefore,

$s_1^2 = C^2 e^2 $

where $e$ is the eccentricity of the given ellipse, $e = \sqrt{1 - \dfrac{b^2}{a^2}} $. Hence,

$c_1^2 = 1 - C^2 e^2 $

Then,

$C^2 - c_1^2 = C^2 (1 + e^2) - 1 $

Substituting this into $(14)$,

$\dfrac{ x_1^2}{a^2 } - z_1^2 \dfrac{ C^2 (1 + e^2) - 1 }{ b^2 C^2 } = 1 \tag{15} $

i.e.

$ \dfrac{x_1^2}{a^2} - z_1^2 \left( \dfrac{1 + e^2}{b^2} - \dfrac{1}{b^2 C^2} \right) = 1 \tag{16}$

Now, from $(11)$ we have

$ x_1^2 (C^2 - s_1^2)^2 = z_1^2 s_1^2 c_1^2 $

but, $s_1^2 = C^2 e^2 $, and $c_1^2 = 1 - C^2 e^2 $

Therefore,

$ x_1^2 C^4 (1 - e^2)^2 = z_1^2 C^2 e^2 (1 - C^2 e^2 ) $

and this leads to

$ C^2 b^2 = \dfrac{b^2 z_1^2 e^2}{ (x1^2 (1 - e^2)^2 + e^4 z1^2 )} $

Substituting this into $(16)$

$ \dfrac{x_1^2}{a^2} - z_1^2 \left( \dfrac{(1 + e^2)}{b^2} + \dfrac{( x_1^2 (1 - e^2)^2 + e^4 z_1^2 )}{b^2 e^2} \right) = 1 \tag{17}$

From which,

$ x_1^2 \left( \dfrac{1}{a^2} + \dfrac{(1 - e^2)^2}{b^2 e^2} \right) - z_1^2 \left( \dfrac{1 + e^2}{b^2} - \dfrac{e^2}{b^2} \right) = 1 \tag{18} $

Re-arranging,

$ x_1^2 \left( \dfrac{b^2 e^2 + a^2 (1 - e^2)^2 }{a^2 b^2 e^2} \right) - \dfrac{z_1^2}{ b^2 } = 1 \tag{19}$

Now $ b^2 e^2 + a^2 (1 - e^2)^2 = b^2 e^2 + a^2 (1 - e^2)\left (\dfrac{b^2}{a^2} )\right) = b^2 (e^2 + 1 - e^2) = b^2 $

Therefore, from this and $(19)$,

$ \dfrac{x_1^2}{a^2 e^2} - \dfrac{z_1^2}{b^2} = 1$

But $a^2 e^2 = a^2 - b^2 $, hence the final equation is

$ \dfrac{x_1^2}{a^2 - b^2} - \dfrac{z_1^2}{b^2} = 1 \tag{20} $

This equation is mentioned here in an answer by intelligenti pauca to a similar question.

So to build such a cone, start by selecting $x_1, z_1$ a point on the hyperbola $(20)$, then compute the cosine of the semi-vertical angle $\phi$, from

$ C^2 = \dfrac{ z_1^2 e^2}{x_1^2 (1 - e^2)^2 + e^4 z_1^2} $

As derived above. Next compute $s_1, c_1$ from

$s_1 = \pm C e$

$c_1 = \pm \sqrt{1 - s_1^2} $

Now the cone is specified (there are four of them) is specified by equations $(4)$ and $(5)$.