8. Let $x \geqslant y>0$, if there exist real numbers $a, b$ satisfying $0 \leqslant a \leqslant x, 0 \leqslant b \leqslant y$, and $(x-a)^{2}+(y-b)^{2}=x^{2}+b^{2}=y^{2}+a^{2}$, then the maximum value of $\frac{x}{y}$ is
A. $\frac{2 \sqrt{3}}{3}$
B. $\sqrt{2}$
C. $\frac{\sqrt{6}}{2}$
D. 1
Answer
As shown in the figure, $\triangle O M N$ is an equilateral triangle. Let $\angle B O N=\alpha$, then we have $O N=\frac{y}{\cos \alpha}, O M=\frac{x}{\cos \left(30^{\circ}-\alpha\right)}$,
Thus, $O M=O N \Rightarrow \frac{x}{\cos \left(30^{\circ}-\alpha\right)}=\frac{y}{\cos \alpha}$ $\Rightarrow \frac{x}{y}=\frac{\cos \left(30^{\circ}-\alpha\right)}{\cos \alpha}=\frac{\sqrt{3}}{2}+\frac{1}{2} \tan \alpha$ $\leqslant \frac{\sqrt{3}}{2}+\frac{1}{2} \cdot \frac{\sqrt{3}}{3}=\frac{2 \sqrt{3}}{3}$, equality holds when $\alpha=30^{\circ}$. Therefore, the answer is $A$.