8. Given the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

with four vertices $A, B, C, D$. If the radius of the inscribed circle of the rhombus $A B C D$ is equal to $\frac{\sqrt{6}}{6}$ of the focal distance of the ellipse, then its eccentricity is $\qquad$.
8. $\frac{\sqrt{2}}{2}$.

From the given, we have
$$
\begin{array}{l}
a b=\frac{\sqrt{6}}{6} \times 2 c \sqrt{a^{2}+b^{2}} \\
\Rightarrow e=\frac{c}{a}=\frac{\sqrt{2}}{2} .
\end{array}
$$