# How many points are needed to uniquely define an ellipsoid centered on the origin?

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...

• No, you cannot identify the ellipsoid from its shadow in the x-y plane. Many ellipsoids tilted at different angles can intersect the x-y plane in the same ellipse. Or consider an ellipsoid with varying diameter on the z axis but fixed diameters on the x and y axes. – Paul Sep 26 '19 at 14:17
• The intersection of the ellipsoid with the $xy$-plane is an ellipse centered at origin. knowing 3 generic points of it is enough to uniquely determine that ellipse. The other 3 points won't give you more information. Another way to interpret this is no matter how many points you know on that ellipse, you only have informations about three of the unknowns ($a, b, h$). – achille hui Sep 26 '19 at 15:56
• @Paul please see my edit. – Armadillo Sep 26 '19 at 16:03
• @achillehui agreed. How would one say in a mathematical way that 6 points are needed but they must provide information for the 6 unknowns ($a,b,c,f,g,h$)? – Armadillo Sep 26 '19 at 17:45
• To make this rigorous, the way I know is probably an overkill for your problem. The basic idea is re-express everything in terms of linear algebra. For each point $p = (x,y,z)$, let $$\tilde{p} = (x^2,y^2,z^2,yz,zx,xy,1)$$ For each conic $C : ax^2 + by^2 + c z^2 + 2fzx + 2gzx + 2hxy = 1$, let $$\tilde{C} = (a,b,c,2f,2g,2h,-1) \in \mathbb{R}^7$$ The condition that $p$ lies on $C$ becomes $\tilde{p} \cdot \tilde{C} = 0$. This is equivalent to saying $\tilde{C}$ lies on a hyperplane orthogonal to $\tilde{p}$. Let's call this hyperplane $\mathcal{H}_p$. – achille hui Sep 26 '19 at 18:32

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 $$0$$.