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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?: $$\tag{3} al_1^2+bl_2^2+cl_3^2+2fl_2l_3+2gl_1l_3+2hl_1l_2$$

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    $\begingroup$ Equations (1) and (2) are equivalent, so they are satisfied by the same $(x,y,z)$ on the surface of the ellipsoid. $\endgroup$ – GReyes Jan 3 '20 at 19:58
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    $\begingroup$ Equation $(2)$ is still an equation of the ellipsoid, it is not scaling it (turning it into a sphere) at all. Take a simple special case where $f, g, h=0$: you end up with $ax^2+by^2+cz^2=1$, i.e. $(x\sqrt{a})^2+(y\sqrt{b})^2+(z\sqrt{c})^2=1$. The point $(x\sqrt{a}, y\sqrt{b}, z\sqrt{c})$ now does lie on a sphere with radius $1$, so it looks like you would like to label $l_1=x\sqrt{a}, l_2=y\sqrt{b}, l_3=z\sqrt{c}$. $\endgroup$ – Stinking Bishop Jan 3 '20 at 20:05
  • $\begingroup$ To add to the previous comment, I am not sure if you call the "position vector" the vector $(x, y, z)$ scaled to magnitude $1$? It is, of course, $\left(\frac{x}{\sqrt{x^2+y^2+z^2}}, \frac{y}{\sqrt{x^2+y^2+z^2}}, \frac{z}{\sqrt{x^2+y^2+z^2}}\right)$, and is, in general, not the same as $(x\sqrt{a}, y\sqrt{b}, z\sqrt{c})$. $\endgroup$ – Stinking Bishop Jan 3 '20 at 20:17
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    $\begingroup$ Multiplying both sides of an equation by the same nonzero constant doesn’t change its solution set. $\endgroup$ – amd Jan 3 '20 at 20:33
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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

$$ \lambda_1 w_1^2+\lambda_2 w_2^2+\lambda_3 w_3^2 = 1 $$

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