8. If $\log _{4}(x+2 y)+\log _{4}(x-2 y)=1$, find the minimum value of $|x|-|y|$.

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8. The original conditions are $\left\{\begin{array}{l}x+2 y>0, \\ x-2 y>0, \\ (x+2 y)(x-2 y)=4\end{array}\right.$, which can be rewritten as $\left\{\begin{array}{l}x>2|y| \geqslant 0 \\ x^{2}-4 y^{2}=4\end{array}\right.$.

By symmetry, we only consider $y \geqslant 0$. Let $x-y=u$, and substitute into $x^{2}-4 y^{2}=4$, we get $3 y^{2}-2 u y+\left(4-u^{2}\right)=0$. $\Delta=16\left(u^{2}-3\right) \geqslant 0$, hence $u \geqslant \sqrt{3}$. And when $x=\frac{4}{3} \sqrt{3}, y=\frac{\sqrt{3}}{3}$, the equality holds.