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 .

Let $\boldsymbol{a}=\left(\sqrt{\frac{x^{2}}{2 y}}, \sqrt{\frac{4 y^{2}}{x}}\right), \boldsymbol{b}=(\sqrt{2 y}, \sqrt{x})$.
Then $|\boldsymbol{a}|=\sqrt{\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}},|\boldsymbol{b}|=\sqrt{2 y+x}=\sqrt{2}$, $\boldsymbol{a} \cdot \boldsymbol{b}=x+2 y=2$.
By $|\boldsymbol{a}||\boldsymbol{b}| \geqslant \boldsymbol{a} \cdot \boldsymbol{b} \Rightarrow \frac{x^{2}}{2 y}+\frac{4 y^{2}}{x} \geqslant 2$, equality holds if and only if $\boldsymbol{a}$ and $\boldsymbol{b}$ are in the same direction.
Therefore, when $x=2 y=1$, $\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}$ achieves its minimum value of 2.