An ellipsoid centered at the origin can be defined by the solutions to
$$ \mathbf{x}^\text{T} A \mathbf{x} = 1 $$ where $A$ is symmetric and positive-definite. The eigenvectors of $A$ define the principal axes and the eigenvalues are the square reciprocal semi-major axes. Call such an $A$ a shape tensor of the ellipsoid $\mathcal{E}$ it defines.
Given a shape tensor $A$, I need to obtain a discretization of $\mathcal{E}$. This discretization must be in the form of a set $\{\mathbf{v}_i\}$ of vectors defined with respect to the coordinate axes. One method of obtaining such a discretization is to first obtain a discretization $\{\mathbf{w}_i\}$ of an identical ellipsoid aligned with the coordinate axes, using a parameterization $(u,v)$ in spherical coordinates and the relations: $$ \begin{align} x&=a\,\cos u\cos v,\\ y&=b\,\cos u\sin v,\\ z&=c\,\sin u,\end{align}\,\! $$
(where $a,b,c$ are the semi-major axes), then compute the Euler angles $\theta$ and $\phi$ of the eigenvectors of $A$, and finally apply a rotation defined by these angles to $\{\mathbf{w}_i\}$.
I strongly suspect this is not necessary. I cannot shake the feeling that I am doing something only to then undo it.
Is there a faster (with respect to numerical computation) way to get this discretization?
Summary: Given the semi-major axes and the directions of the principal axes of an ellipsoid (i.e., $A$), how can I obtain a discretization of this ellipsoid?
"I cannot shake the feeling that I am doing something only to then undo it."
Because $A$ is symmetric there exists an orthogonal set of eigenvectors of $A$. Normalizing these produces an orthonormal basis, with associated rotation matrix consisting of these three eigenvectors as columns. Furthermore, the eigenvectors associated with different eigenvalues will always be orthogonal. Thus, given the eigenvectors of $A$ with the additional restriction that they be orthogonal, the problem is solved: the rotation matrix is trivially constructed. This leaves only the issue of orthogonality in the case of repeated eigenvalues, which depends on the algorithm used to generate the eigenvectors; a question for Stack Overflow.