Formula for an Ellipsoid 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!
 A: 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 $\det M>0$.
A: 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.
