On conics curves and increasing unions of ellipses It is easy to see that the epigraph of a parabola, i.e. the set
$
\\{(x,y)\in \mathbb R^2, y> x^2\\}
$
 is a countable increasing union of ellipses in the sense that
$$
\\{(x,y)\in \mathbb R^2, y> x^2\\}=\cup_{k\ge 1}\\{(x,y)\in \mathbb R^2, y\ge x^2+k^{-2} y^2\\}.
$$
On the other hand, I believe that the epigraph of an hyperbola defined as 
$$\\{(x,y)\in \mathbb R^2, y> \sqrt{x^2+1}\\}$$
cannot be a countable increasing union of ellipses. However, the proof that I have of this fact is very indirect and is using some rather complicated stuff about singular integrals. It is quite likely that there is a simple planar Euclidean geometry argument to support the above claim.
Maybe a simple argument about the eccentricity of the hyperbola (which is $>1$)
can prevent that it is the union of ellipses (whose eccentricity is $<1$).
 A: Here is a proof of this fact. Suppose by contradiction there is such a family of ellipses. Each of them is given by an inequality $P_n(x,y)\le 0$, where $P_n$ is a degree two polynomial, $P_n=Q_n+L_n+c_n$. Here $c_n$ is a constant, $L_n$ is the linear part, and $Q_n $ is a positive definite quadratic form. 
Let's normalise each $P_n$ so that the sum of squares of its coefficients is $1$. Up to passing to a convergent subsequence we can assume that $P_n$ converges to $P$. 
We can say the following about $P$. 1) $P$ is non-positive on all the ellipses $E_n$, in particular on their union and so on the epigraph of the hyperbola. 2) $P$ is non-negative on the complement to the epigraph of the hyperbola. 3) The quadratic part $Q$ of $P$ is either a) positive definite, or b) positive semindefinite. 
In case a) the set $P\ge 0$ is an ellipse, so it is smaller than the epigraph of the hyperbola. In case b) the set $P\ge 0$ is an interior of a parabola (not an exterior thanks to 2) ). So again, it is smaller than the epigraph of the hyperbola. Both situations contradict 1).
A: Thanks, that's great. Let me write my comment as an answer since I have only 5 minutes to write down a comment, which is a bit short.
Some minor remarks on your two last paragraphs. Let's call $\mathcal H=\{(x,y)\in \mathbb R^2, y> \sqrt{x^2+1}\}$ and let us assume that 
$$
\mathcal H=\cup_n\{P_n(x,y)\le 0\}\quad\text{increasing union}.
$$
 Indeed we have $\mathcal H\subset\{P\le 0\}$ for your reason (1) and also $\mathcal H^c\subset\{P\ge 0\}$ since the $P_n$ are $> 0$ on $\mathcal H^c$.
The quadratic part $Q$ of $P$ is non-negative: if rank $Q=2$, that would force the inclusion of the unbounded $\mathcal H$ in the compact $\{P\le 0\}$. If rank $Q$=1,
that
 would force the inclusion of $\mathcal H^c$ in the parabola $\{P\ge 0\}$ which is also impossible, but deserves a short proof. If $Q=0$ then $P$ is a non-zero affine function vanishing on the boundary of $\mathcal H$, which is also impossible.
