1
$\begingroup$

Hi fellow mathematicians!

I am having a hard time figuring out how an ellipsoid is globally defined.

I know that a local formula for an ellipsoid is:

$(\frac{x}{A})^2 + (\frac{y}{B})^2 + (\frac{z}{C})^2 - 1 = 0$.

However, with this formula I can only place it on the coordinate beginning and it would be aligned with axis. I would like to be able to place it anywhere with any rotation I could think of.

I have found a formula that should globally describe an ellipsoid:

$Ax^2 + By^2 + Cz^2 + Dyz + Ezx + Fxy + Gx + Hy + Jz + K = 0$.

Unfortunately, I have no idea how we got such a formula from the local one and I do not know how to use it. It would be really, really appreciated if someone could explain it to me.

Notation: Capital letters are constants and $x$, $y$ and $z$ are coordinates.

Thanks a lot!

$\endgroup$
0
$\begingroup$

What you have there is the general equation of a quadric surface. It could be an ellipsoid, a hyperboloid of one or two sheets, a paraboloid etc. It is an ellipsoid if the matrix $$M=\pmatrix{A&F/2&E/2\\F/2&B&D/2\\E/2&D/2&C}$$ is positive definite, that is $v^tMv>0$ for any nonzero real vector $v$. One condition that is equivalent to the positive definiteness of $M$ is that $A>0$, $AB>F^2/4$ and $\det M>0$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a lot for a quick response! What happened with the constants $G$, $H$, $J$ and $K$? Where do they come from? I guess you have a typo in the second row, third column. Shouldn't it be $D/2$? I tried to plot an ellipsoid as: $5x^2 + 5y^2 + 10 z^2 + 8yz + 6zx + 2xy = 0$ Which should be OK with the conditions given for matrix $M$, but the plot didn't show a surface. Any idea what the problem could be? $\endgroup$ – Leta May 12 '17 at 18:10
  • $\begingroup$ Silly me, the input I used is just plain wrong... $\endgroup$ – Leta May 12 '17 at 18:22
  • $\begingroup$ The constants $G$, $H$, $J$ help determine the centre of the ellipsoid. There should also be a condition on $K$ which I don't really want to write down: if $K$ is big the solution set is empty, like for the "sphere" $x^2+y^2+z^2+1=0$. $\endgroup$ – Angina Seng May 12 '17 at 18:39
0
$\begingroup$

A general technique you can use (and generalize) is the following:

Let $T : \mathbb{R}^3 \to \mathbb{R}^3$ be an invertible affine transformation, so $T(x,y,z) = A (x,y,z) + v$, where $A$ is an invertible linear transformation and $v$ is a vector.

Let $S \subset \mathbb{R}^3$ be the zero set of some polynomial(s) $f$. Then the shape $T(S)$ is the zero set of $f \circ T^{-1}$. Now you can compute the polynomial $f \circ T^{-1}$ to get your formula. (To do this, it might be helpful to think of $T^{-1}$ as a transformation from $\mathbb{R}^3$ with $x,y,z$ coordiantes to another $\mathbb{R}^3$ with $x'$,$y'$ and $z'$ coordinates, and $f$ a function in $x',y',z'$. Then the transformation $T^{-1}$ is expressed by some transformation rules such as $x' = \text{linear}(x,y,z) + a$, and to find the polynomial $f \circ T^{-1}$ you use substitution.)

(The verification is like this : If $T(s) \in T(S)$, then $f \circ T^{-1} (Ts) = f(s) = 0$, and if $f (T^{-1}(w)) = 0$, then $f ( T^{-1}(w) ) = 0$, so $T^{-1}w \in S$, so $w \in T(S)$.)

So in your case, you can take $f$ to be your "local equation" (I don't think this the right terminology)and then take $T$ to be a translation, or a rotation of the axis.

| 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.