It can be seen that the points of tangency of the spheres with $E$ must lie on its major axis due to symmetry. Hence, we consider the two-dimensional cross-section with plane $ABC$. Then the two spheres become the incentre and the excentre of the triangle $ABC$, and we are looking for the ratio of the inradius to the exradius. Let $s, r, r_{a}$ denote the semiperimeter, inradius, and exradius (opposite to $A$ ) of the triangle $ABC$. We know that the area of $ABC$ can be expressed as both $rs$ and $r_{a}(s-|BC|)$, and so $\frac{r}{r_{a}}=\frac{s-|BC|}{s}$. For the given triangle, $s=6$ and $a=4$, so the required ratio is $\frac{1}{3}$. $\qquad$