6. Let $x, y \in \mathbf{R}$, and $\log _{4}(x+2 y)+\log _{4}(x-2 y)=1$, then the minimum value of $x-|y|$ is ( )
A: $\sqrt{3}$
B: 2
C: $\quad 2 \sqrt{3}$
D: 4
Answer $\mathrm{A}$.
Analysis From the problem, we have $x^{2}-4 y^{2}=4$ and $x>2|y|$. Let's assume $x=2 \sec \theta, y=\tan \theta, \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so
$$
x-|y|=\frac{2}{\cos \theta}-\frac{|\sin \theta|}{\cos \theta}
$$

The right side of the equation is an even function of $\theta$, so we can assume
$$
t=\frac{2-\sin \theta}{\cos \theta}, \theta \in\left[0, \frac{\pi}{2}\right) \Rightarrow \sin \theta+t \cos \theta=2
$$

Therefore,
$$
\sqrt{t^{2}+1} \sin (\theta+\varphi)=2 \Rightarrow \frac{2}{\sqrt{t^{2}+1}} \leq 1 \Rightarrow t^{2} \geq 3
$$

Thus, the minimum value $\sqrt{3}$ is achieved if and only if $x=\frac{4}{\sqrt{3}},|y|=\frac{1}{\sqrt{3}}$.