Given two ellipses $E_1,E_2$ of equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=4$$ prove that for all $p\in E_1$ there exist a unique ellipse $F_p$ that meets $E_1$ in $p$ and in one other point, and $E_2$ in two points too.
The question is: is the statement true? If one chooses $a=b=1$, we have two concentric circles. Then choosing $p=(0,1)$, one can consider many circles that meet the smaller circle in $p$ and, e.g., in $(1,0)$, and the greater circle in two other points.
Am I missing something that should restrict the choice range?
I would say you are not missing anything. There is a linear map (a dilation) $\varphi$ that brings the original concentric and coaxial ellipses $E_1,E_2$ into two concentric circles $\varphi(E_1),\varphi(E_2)$. It is enough to use a rotation to find an ellipse $L$ through $\varphi(P)$ that is tangent to $\varphi(E_1)$ at $\varphi(P)$ and is tangent to $\varphi(E_2)$, too. $L$ is just an ellipse with its semi-axis given by the radii of $\varphi(E_1),\varphi(E_2)$. Then $\varphi^{-1}(L)$ is an ellipse fulfilling the wanted constraints.