6. As shown in Figure 1, in a cylindrical ping-pong ball container with a height of 10 and a base radius of 1, a ping-pong ball is attached to both the top and bottom walls of the container. The ping-pong ball on the bottom wall is tangent to the bottom and the side of the container, and the ping-pong ball on the top wall is tangent to the top and the side of the container (the thickness of the container and the ping-pong balls is negligible). A plane is tangent to both ping-pong balls. It is known that the shape formed by the plane intersecting the edge of the container is an ellipse. Then the eccentricity of this ellipse is $\qquad$
6. $\frac{\sqrt{15}}{4}$.
Let the lengths of the semi-major axis and semi-minor axis of the ellipse be $a, b$ respectively. As shown in Figure 8, construct the axial section of the cylinder so that the intersection line of this section with the plane of the ellipse is exactly the major axis of the ellipse. Let the endpoints of the major axis of the ellipse be $A, B$, and the midpoint be $O$. Let the centers of the two spheres be $O_{1}, O_{2}$, and by symmetry, $O$ is the midpoint of $O_{1} O_{2}$. The axial section intersects the two spheres to form two great circles $\odot O_{1}, \odot O_{2}$. Suppose $A B$ is tangent to $\odot O_{1}$ at point $R$, then in the right triangle $\triangle O O_{1} R$, $\sin \angle O_{1} O R=\frac{O_{1} R}{O O_{1}}=\frac{1}{4}$. Draw $O M \perp O_{1} O_{2}$ through point $O$, intersecting the boundary of the axial section at point $M$. It is easy to see that $\angle O_{1} O R=\angle O A M$, thus $O A=\frac{O M}{\sin \angle O A M}=4$, i.e., $a=4$. By symmetry, it is easy to know that the semi-minor axis length of the ellipse is the radius of the base of the cylinder, i.e., $b=1$. Therefore, the semi-focal distance of the ellipse $c=\sqrt{a^{2}-b^{2}}=\sqrt{15}$, so the eccentricity of the ellipse is $\frac{\sqrt{15}}{4}$.