8. Given $x, y>0$, and $x+2 y=2$. Then the minimum value of $\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}$ is . $\qquad$
8. 2 .
$$
\begin{array}{l}
\text { Let } \boldsymbol{a}=\left(\sqrt{\frac{x^{2}}{2 y}}, \sqrt{\frac{4 y^{2}}{x}}\right), \boldsymbol{b}=(\sqrt{2 y}, \sqrt{x}) . \\
\text { Then }|\boldsymbol{a}|=\sqrt{\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}},|\boldsymbol{b}|=\sqrt{2 y+x}=\sqrt{2} \\
\boldsymbol{a} \cdot \boldsymbol{b}=x+2 y=2 .
\end{array}
$$

From $|\boldsymbol{a}||\boldsymbol{b}| \geqslant \boldsymbol{a} \cdot \boldsymbol{b} \Rightarrow \frac{x^{2}}{2 y}+\frac{4 y^{2}}{x} \geqslant 2$, equality holds if and only if $\boldsymbol{a}$ is in the same direction as $\boldsymbol{b}$.
Therefore, when $x=2 y=1$, $\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}$ achieves its minimum value of 2.