An equation of an ellipsoid, when referred to its center as the coordinate system's origin, can be written as $$\tag{1} Ax^2+By^2+Cz^2+2Fyz+2Gxz+2Hxy+D=0$$
As I understand things, the values used for $x, y, z$ would be the Cartesian $x,y,z$ coordinates that describe the position of the points that make up the ellipsoid's surface.
If we were to multiply the coefficients in Eqn(1) by $(-1/D)$ we would then have $$\tag{2} ax^2+by^2+cz^2+2fyz+2gxz+2hxy=1$$
Would it be true that the $x,y,z$ values used in Eqn(2) are not the Cartesian coordinates of the points of the surface? Rather, the values to be used for $x,y,z$ are the components of the unit vector $l_i (l_1,l_2,l_3)$ that describes the orientation of the position vector that represents the position of a point on the ellipsoids surface? (I.e. the position vector is a radius vector $r$ drawn out from the origin in a direction pointing to the particular location on the surface of the ellipsoid).
Thus, we should write?: $$\tag{3} al_1^2+bl_2^2+cl_3^2+2fl_2l_3+2gl_1l_3+2hl_1l_2$$
Given
$$ M = \left( \begin{array}{ccc} a & h & g \\ h & b & f \\ g & f & c \\ \end{array} \right) $$
an hermitian matrix and $X = (x,y,z)^{\dagger}$ we have
$$ X^{\dagger}MX = 1 $$
represents an ellipsoid as long as $\Lambda$ in
$$ M = Q^{-1}\Lambda Q $$
is a diagonal pos1tive matrix.
Here $Q$ is the normalized eigenvectors matrix associated to $M$. Making a coordinates change as $W = (w_1,w_2,w_3)^{\dagger} = Q X$ we have
$$ W^{\dagger}\Lambda W = 1 $$
This new coordinate system has for axes the eigenvectors of $M$ and in the $W$ coordinates the ellipsoid reads
$$ \lambda_1 w_1^2+\lambda_2 w_2^2+\lambda_3 w_3^2 = 1 $$