Check if the union of hyper-ellipses cover the unit sphere Given $N$ hyper ellipses characterized by positive definite matrices $A_1,\dots,A_N\in\mathbb{R}^{n\times n}$ and positive definite constants $\rho_1,\dots,\rho_N$. This is,
$$
\mathcal{E}_i  = \{x: x\in\mathbb{R}^n \text{ and } x^TA_ix\leq \rho_i\}\subset \mathbb{R}^n,\ \ i=1,\dots,N.
$$
How can we check if $S_n\subset\bigcup_{i=1}^N\mathcal{E}_i$, where $S_n$ is the unit sphere in $\mathbb{R}^n$?
Initially I thought that since $S_n$ is convex and all ellipses are centered at the origin then $S_n\subset\bigcup_{i=1}^N\mathcal{E}_i$ only if $S_n\subset\bigcap_{i=1}^N\mathcal{E}_i$. Then I could check if $S_n$ was inside all $\mathcal{E}_i$, which is easy. However, this is not true in general, and $S_n\subset\bigcap_{i=1}^N\mathcal{E}_i$ is only a sufficient condition.
I am more interested in cases such as

I think, it has to do with checking if the distance between the origin, and the points of intersection between the boundaries of the ellipses is greater than 1, But I dont know how this would apply in any case different than $\mathbb{R}^2$, or in the case the boundaries does not intersect.
Do you have any idea?
 A: I reformulate the problem as follows: taking $B_i = A_i/\rho_i$, we want to check whether
$$
x^Tx \leq 1 \implies \exists i : x^TB_i x \leq 1.
$$
In the $2$ dimensional case, one approach is to compute $S^1 \cap \mathcal E^i$ for all $i$, where $S^1 = \{x \in \Bbb R^2 : \|x\| = 1\}$. $\bigcup \mathcal E^i$ contains the unit ball if and only if the union of these intersections is $S^1$.
We can do this systematically find these intersectoins as follows: let $\mathcal E = \{x : x^TBx \leq 1\}$. Let $\lambda_1 \leq \lambda_2$ denote the eigenvalues of $B$, let $v_1$ denote the eigenvector associated with $\lambda_1$, and let $R$ denote the clockwise rotation by $90^\circ$.
If $\lambda_1 \geq 1$, then $\mathcal E$ does not contain $S^1$. If $\lambda_2 \leq 1$, then $\mathcal E$ contains all of $S^1$.
In the remaining case, we note that if we take $x = \cos \theta v_1 + \sin\theta Rv_1$, then we have
$$
x^TBx = \lambda_1 \cos^2\theta + \lambda_2 \sin^2\theta = \lambda_1 + (\lambda_2 - \lambda_1)\sin^2 \theta.
$$
So, we have
$$
x^TBx = 1 \implies (\lambda_2 - \lambda_1)\sin^2 \theta = 1 - \lambda_1 \implies\\
\sin\theta =  \pm\sqrt{\frac{1 - \lambda_1}{\lambda_2 - \lambda_1}}
$$
The four solutions to this equation break the circle $S^1 = \{\cos\theta v_1 + \sin \theta Rv_1 : \theta \in [0,2\pi)\}$ into four arcs. The arc that contains $v_1$ and the arc that contains $-v_1$ lies within $\mathcal E_2$, and the remaining open arcs do not.
