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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!

Leta
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2 Answers2

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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 $\det M>0$.

Angina Seng
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  • 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?

     – Leta May 12 '17 at 18:10
  • Silly me, the input I used is just plain wrong... – Leta May 12 '17 at 18:22
  • 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" $x^2+y^2+z^2+1=0$. – Angina Seng May 12 '17 at 18:39
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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.

Elle Najt
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