If I have 6 points that describes an ellipse-of-intersection of the x-y plane intersecting an ellipsoid centered at the origin, and the ellipsoid is not necessarily axis-aligned, do these six points have enough information to uniquely define the ellipsoid?
I have read that the equation for an ellipsoid centered on the origin can be written as: $$ax^2+by^2+cz^2+2fyz+2gzx+2hxy-1=0$$ And that since this equation has six unknowns, $a,b,c,f,g,$ and $h$, then, at the minimum, six general points on the ellipsoid fix them.
If this true then the 6 points I have describing the ellipse-of-intersection should be enough to determine the six unknowns ($a,b,c,f,g,h$) of the ellipsoid. Is that correct?
Edit: If the statement is not correct, what would be the more correct statement as opposed to the statement previously made, "since this equation has six unknowns, a, b, c, f, g, and h, then, at the minimum, six general points on the ellipsoid fix them." ?
I'd imagine it would be something along the lines of, "a minimum of six points are needed, were there at least one point that provides data to the 2fyz, 2gzx, and 2hxy terms," but I am not sure...
Putting as equation $a_1x^2+a_2y^2+a_3z^2+a_4yz+a_5zx+a_6xy=1$ the six points $(x_i,y_i)$ give the linear system
$$\begin{matrix} n_{11}a_1+n_{12}a_2+\cdots+n_{16}a_6=1\\ n_{21}a_1+n_{22}a_2+\cdots+n_{26}a_6=1\\ n_{31}a_1+n_{32}a_2+\cdots+n_{36}a_6=1\\ n_{41}a_1+n_{42}a_2+\cdots+n_{46}a_6=1\\ n_{51}a_1+n_{52}a_2+\cdots+n_{56}a_6=1\\ n_{61}a_1+n_{62}a_2+\cdots+n_{66}a_6=1\end{matrix}$$ The condition for this system have a unique solution is that the determinant of the matrix of the $n_{ij}$ where $1\le i,j\le6$ be distinct of $0$.
Now, clearly not any arbitrary set of six points of the ellipsoide satisfies this condition.(Can you choose a such set? and also a set of six point determining a unique ellipsoide?