Proof for defining a unique ellipsoid

Question

Is it possible to show that, of the 6 points needed to uniquely define an ellipsoid centered at the origin, no point can be co-linear with another point and the origin, and no more than 3 points can be co-planar?

Answer 1

Ellipsoid centered at 0 is essentially a symmetric bilinear form $A$, so $x^TAx=1$ on points of ellipsoid. Symmetric form of dimension 3 has 6 degrees of freedoms (6 unknowns), so six points give six linear equations of form $x^TAx=1$ which is in general is enough to solve for 6 varialbles.

However, if two points $x$ and $y$ a collinear with the origin, then $y=\alpha x$, and equations of those points: $x^TAx=1$ and $y^TAy=\alpha^2 x^TAx=1$. So either they are incompatible ($\alpha^2\neq 1$) or they are essentially the same equation $\alpha^2=1$. In first case, there is no ellipsoid, in second, there are infinitely many ellipsoids.

If three points $x,y,z$ are collinear, then $z=\alpha x+\beta y$. So again equations: $x^TAx=1$, $y^TAy=1$ and $z^TAz=\alpha^2(x^TAx)+\beta^2(y^TAy)=1$ are dependent.

You can apply the same logic for 4 coplanar points.