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5. (5 分) 已知 $M\left(x_{0}, y_{0}\right)$ 是双曲线 $C: \frac{x^{2}}{2}-y^{2}=1$ 上的一点, $F_{1}, F_{2}$ 是 $C$ 的左 、右两个焦点, 若 $\overrightarrow{M F_{1}} \cdot \overrightarrow{\mathrm{MF}_{2}}<0$, 则 $\mathrm{y}_{0}$ 的取值范围是( )

Gaokao · Math · previous-year question

  1. A.$\left(\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right)$correct
  2. B.$\left(-\frac{\sqrt{3}}{6}, \frac{\sqrt{3}}{6}\right)$
  3. C.$\left(-\frac{2 \sqrt{2}}{3}, \frac{2 \sqrt{2}}{3}\right)$
  4. D.$\left(-\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}\right)$

Answer

A. $\left(\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right)$

Explanation

解:由题意, $\overrightarrow{M_{1}} \cdot \overrightarrow{M_{2}}=\left(-\sqrt{3}-x_{0},-y_{0}\right) \cdot\left(\sqrt{3}-x_{0},-y_{0}\right)$ $$ =\mathrm{x}_{0}^{2}-3+\mathrm{y}_{0}^{2}=3 \mathrm{y}_{0}^{2}-1<0 $$ 所以 $-\frac{\sqrt{3}}{3}<y_{0}<\frac{\sqrt{3}}{3}$. 故选: $A$.

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