Normalizing (scaling) an equation of an ellipsoid, what becomes of the $x,y,z$ terms?

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

• Equations (1) and (2) are equivalent, so they are satisfied by the same $(x,y,z)$ on the surface of the ellipsoid. – GReyes Jan 3 '20 at 19:58
• 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 $l_1=x\sqrt{a}, l_2=y\sqrt{b}, l_3=z\sqrt{c}$. – Stinking Bishop Jan 3 '20 at 20:05
• 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 $(x\sqrt{a}, y\sqrt{b}, z\sqrt{c})$. – Stinking Bishop Jan 3 '20 at 20:17
• Multiplying both sides of an equation by the same nonzero constant doesn’t change its solution set. – amd Jan 3 '20 at 20:33

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

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