8. For the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the area of the section through the diagonal $B D_{1}$ is $S$. Let $S_{\max }$ and $S_{\text {min }}$ be the maximum and minimum values of $S$, respectively. Then the value of $\frac{S_{\max }}{S_{\text {min }}}$ is ( ).
A. $\frac{\sqrt{3}}{2}$
B. $\frac{\sqrt{6}}{2}$
C. $\frac{2 \sqrt{3}}{3}$
D. $\frac{2 \sqrt{6}}{3}$
8. C. As shown in the figure, let the section through $B D_{1}$ be the parallelogram $B E D_{1} E_{1}$, then $S=2 S_{\triangle E D_{1} E_{1}}=h \cdot B D_{1}$, where $h$ is the distance from $E_{1}$ to $B D_{1}$. It is easy to see that $h$ is minimized when it is the distance $h^{\prime}$ between the skew lines $B D_{1}$ and $B_{1} C_{1}$. At this time, $h=h^{\prime}=\frac{a}{\sqrt{2}}(a$ is the edge length of the cube), thus
$$
\begin{array}{c}
S_{\text {min }}=\frac{a}{\sqrt{2}} \cdot \sqrt{3} a=\frac{\sqrt{6}}{2} a^{2} . \\
\text { Also } S_{\max }=S_{\square B B_{1} D_{1} D}=\sqrt{2} a^{2},
\end{array}
$$

Therefore, $\frac{S_{\max }}{S_{\min }}=\frac{2 \sqrt{3}}{3}$.