Gaokao Math Practice Question
已知双曲线 $C: \frac{x^{2}}{3}-y^{2}=1, O$ 为坐标原点, $F$ 为 $C$ 的右焦点, 过 $F$ 的直线与 $C$ 的两条渐近线的交点分别为 $M, N$. 若 $\triangle O M N$ 为直角三角形, 则 $|\mathrm{MN}|=(\qquad)$
- A.$\frac{3}{2}$
- B.3
- C.$2 \sqrt{3}$
- D.4
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