Gaokao Math Practice Question
11. (5 分) 已知椭圆 C: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ 的左、右顶点分别为 $A_{1}, A_{2}$, 且以线段 $A_{1} A_{2}$ 为直径的圆与直线 $b x-a y+2 a b=0$ 相切, 则 $C$ 的离心率为
- A.$\frac{\sqrt{6}}{3}$
- B.$\frac{\sqrt{3}}{3}$
- C.$\frac{\sqrt{2}}{3}$
- D.$\frac{1}{3}$
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