How to determine an ellipse in 3D?

Question

I know that conic sections are formed by the intersection of a cone and a plane. For example, let $C$ be the cone and $P$ the plane,

$$C=\{(x,y,z) \in \Bbb R^3 \mid x^2+y^2=z^2\}$$

$$P=\{(x,y,z) \in \Bbb R^3 \mid x+2y+5z=5\}$$

I know that their intersection gives an ellipse. But are there any other ways to determinate this ellipse? (Any other way than "Cone-plan-intersection determination". Perhaps a Cartesian equation?

You can eliminate $z$ and recover an equation depending only on $x$ and $y$. — user170231, Jun 21 '23 at 15:48

Intelligenti pauca · Accepted Answer · 2023-06-21T18:18:46.400

Another way of representing an ellipse is by a parametric equation: $$ E(t)=M+\vec a\cos t+\vec b\sin t, $$ where $M$ is the centre of the ellipse, vectors $\vec a$ and $\vec b$ are the semi-major axis and semi-minor axis. Let's find it in this case.

The major axis of the ellipse lies on a plane $Q$ passing through the axis of the cone and perpendicular to the plane of the ellipse. In the case at hand, the equation of $Q$ is $$ Q=\{(x,y,z) \in \Bbb R^3 \mid y=2x\}. $$ The intersections between $Q$, $P$ and $C$ are then the vertices of the ellipse (endpoints of its major axis), which can thus be found $$ A=\left({-1-\sqrt5\over4},{-1-\sqrt5\over2},{5+\sqrt5\over4} \right), \quad B=\left({-1+\sqrt5\over4},{-1+\sqrt5\over2},{5-\sqrt5\over4} \right). $$ The centre $M$ of the ellipse is the midpoint of $AB$: $$ M={1\over4}(-1,-2,5) $$ and vector $\vec a=A-M$ is then: $$ \vec a={\sqrt5\over4}(-1,-2,1). $$ To find vector $\vec b$ note first of all that it is perpendicular to $\vec a$ and to the axis of the cone, i.e. $\vec b\propto (2,-1,0)$. Its module can be found (as explained here) from $$ b^2=\overline{OA}\cdot \overline{OB}\cdot\sin^2 u, $$ where $O=(0,0,0)$ is the vertex of the cone and $u=45°$ is its semi-aperture. This gives: $b^2=5/4$, hence: $$ \vec b={1\over2}(2,-1,0). $$ A parametric equation of the ellipse can then be written as explained above.

Thanks a lot! One more question if you still have time: Is there any way to find the foci and excentricity? — máthēma, Jun 21 '23 at 19:22
@máthēma Of course: foci lie on segment $AB$, at a distance $c=\sqrt{a^2-b^2}=\sqrt{10}/4$ from centre $M$. And eccentricity is $c/a=1/\sqrt3$. — Intelligenti pauca, Jun 21 '23 at 19:57
Thank you! One last question: Is there any way to find the coordinates of the foci? Instead of the distance from the origin. ☺️ — máthēma, Jun 22 '23 at 16:41

score 2 · Answer 2 · answered Jun 24 '23 at 11:33

Generic equation

$$E(t)=M+\vec a\cos t+\vec b\sin t$$

given in the answer of @Intelligenti Pauca can be transformed into :

$$E(t)=\pmatrix{m_1\\m_2\\m_3}+\pmatrix{a_1\\a_2\\a_3}\cos t+\pmatrix{b_1\\b_2\\b_3}\sin t=\pmatrix{m_1+a_1 \cos t +b_1 \sin t\\m_2+a_2 \cos t +b_2 \sin t\\m_3+a_3 \cos t +b_3 \sin t}$$

$$E(t)=\pmatrix{m_1+c_1 \cos(t-\alpha_1)\\m_2+c_2 \cos(t-\alpha_2)\\m_3+c_3 \cos(t-\alpha_3)}$$

with $c_k=\sqrt{a_k^2+b_k^2}$ and $\alpha_k=\tan^{-1}\tfrac{b_k}{a_k}$ if we want to connect these values with the initial values or with random values for the $m_k,c_k,a_k$.

For certain rules on the $m_k,c_k,a_k$, one gets interesting families.

clear all;close all;hold on;
n=50;t=0:0.01:2*pi;
for k=-n:n
   c1=20+k/2;c2=20+k;c3=20+2*k;
   inc=2*pi/n;
   a1=c2*inc;a2=c1*inc;a3=0;
   plot3(c1*cos(t-a1),c2*cos(t-a2),c3*cos(t-a3),'linewidth',5);view([-30,45]);
end;

How to determine an ellipse in 3D?

2 Answers2