15
$\begingroup$

I have a question concerning the formulation of an (3D) ellipsoid. The most common definition for an ellipsoid seems to be:

$E = \{ x=\left( x_1, \dots x_n \right)^T \in R^n: \sum_{i=1}^n \left( \frac{x_i}{r_i} \right)^2 = 1 \}$

based on the different radii $r_i$. However for an ellipse (2D) I also found the following definition:

$E = \{ x \in R^2: \| x - f_1 \| + \| x - f_2 \| = c \}$

which is based on the two focal points $f_1$ and $f_2$ of an ellipse. Now my question is, if there exist such kind of definitions also for ellipsoid in 3D or higher? How many focal points can an ellipsoid in 3D or higher actually have? I am especially interested in ellipsoids where all radii $r_1, \dots, r_n$ can be different (like a Scalene or triaxial ellipsoids).

Many thanks in advance

$\endgroup$
  • $\begingroup$ We should investigate if any biaxial (that is, by a plane containing any two ellipsoid axes) section of an ellipsoid gives an ellipse. Both answers give strange results, like having ellipse with four foci or with no foci at all. $\endgroup$ – mbaitoff Feb 1 '11 at 11:17
  • 1
    $\begingroup$ If I remember correctly, the analogue of the pair of focal points for an ellipsoid in 3D are a pair of curves, namely an ellipse and a hyperbola (in two orthogonal planes). Unless someone else gives an answer, I will try to look into this and return later. $\endgroup$ – Hans Lundmark Feb 1 '11 at 11:20
  • 4
    $\begingroup$ Hmm, maybe this is it: books.google.com/… $\endgroup$ – Hans Lundmark Feb 1 '11 at 11:22
3
$\begingroup$

For confocal central quadrics,

$$\frac{x^2}{a^2+s}+\frac{y^2}{b^2+s}+\frac{z^2}{c^2+s}=1$$

where $a>b>c\geq 0$, there're one focal ellipse namely,

$$\left \{ \begin{array}{rcl} \displaystyle \frac{x^2}{a^2-c^2}+\frac{y^2}{b^2-c^2} &=& 1 \\ z &=& 0 \end{array} \right.$$

and one focal hyperbola namely,

$$\left \{ \begin{array}{rcl} \displaystyle \frac{x^2}{a^2-b^2}-\frac{z^2}{b^2-c^2} &=& 1 \\ y &=& 0 \end{array} \right.$$

For confocal paraboloids,

$$\frac{x^2}{a^2+s}+\frac{y^2}{b^2+s}=z+\frac{s}{4}$$

where $a>b\geq 0$, there're two focal parabolae namely,

$$\left \{ \begin{array}{rcl} \displaystyle \frac{x^2}{a^2-b^2} &=& \displaystyle z-\frac{b^2}{4} \\ y &=& 0 \end{array} \right.$$

and

$$\left \{ \begin{array}{rcl} \displaystyle \frac{y^2}{b^2-a^2} &=& \displaystyle z-\frac{a^2}{4} \\ x &=& 0 \end{array} \right.$$

There's so-called thread construction, please see p.19 of this onwards.

Note that the two focal conics are called conjugate conics and their eccentricities are reciprocal to each other.

Any point on the conjugate conics is the radiant point of the enveloping cone for that conic and of its Dandelin sphere.

enter image description here

P.S.:

The lines of curvature of an ellipsoid actually are the geodesic ellipses and geodesic hyperbolae from two adjacent umbilical points.

See more details here.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Hint:

By symmetry, a quadric with two focal points is an ellipsoid of revolution and can't be more.

In 2D, a conic is defined by $5$ independent parameters, which can be the coordinates of two foci and the sum of distances.

A general quadric needs $9$ parameters, which cannot equal $3n+1$. (Not a proof but a serious clue; similarly, the reduced conic has only $3$ parameters, which cannot equal $3n-5$, where $6$ DOFs have been absorbed by a similarity transform.)

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

I recently proved the two following related results:

THEOREM 1. Let $O$ be the origin point of $\mathbf{R}^n$ and let $A_1, A_2, F_1$ and $F_2$ be four points located on the main axis of $\mathbf{R}^n$, with $A_1$ and $F_1$ on the positive side and $A_2$ and $F_2$ on the negative side. Suppose that $a>b$, and $\phi$ are three constants, with \begin{align} \label{OA1_A2O_a} OA_1 = A_2O &= a \quad OF_1=F_2O = \phi \\ \label{a2_b2phi2} \text{and}\quad a^2&= b^2+\phi^2. \end{align} Let $\mathbf{C}\subset \mathbf{R}^n$ be a set of points, with $A_1, A_2\in \mathbf{C}$. Suppose that $\mathbf{C}$ is an $n$ dimensional centered ellipsoid defined by the equation: \begin{equation} \hspace{5cm}\frac{x_1^2}{a^2} + \frac {x_2^2}{b_1^2}+ \ldots+ \frac{x_n^2}{b_{n-1}^2}=1.\hspace{6cm} (*) \end{equation} Then, the two following conditions are equivalent:

(i) We have $b_1=b_2=\ldots=b_{n-1}=b$ in Equation $(*)$. That is, for any ${\bf x} = \{x_1,\ldots,x_n\}\in \mathbf{C}$, we have \begin{align} \frac{x_1^2}{a^2} + \frac {x_2^2}{b^2}+ \ldots+ \frac{x_n^2}{b^2}=1. \end{align}

(ii) The set $\mathbf{C}$ satisfies the Two Foci Property. That is, for any point $\bf x \in \mathbf{C}$, the sum of the two distances between $\bf x$ and the foci $F_1$ and $F_2$ is the constant $2a$.

DEFINITION. An $n$-dimensional, centered, ellipsoid is called REGULAR if it satisfies either of the two above conditions.

THEOREM 2. The non-empty intersection of a regular centered $n$-dimensional ellipsoid with a plane perpendicular to the main axis of the ellipsoid is either a point or a circle.

CONJECTURE. The non-empty intersection of a regular centered $n$-dimensional ellipsoid with a plane perpendicular to any minor axis of the ellipsoid is either a point or an ellipse.


These results may very well be already know. If they are not, what would be a suitable journal?

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.