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10. (5 分) 设 $F$ 为抛物线 $C: y^{2}=3 x$ 的焦点, 过 $F$ 且倾斜角为 $30^{\circ}$ 的直线交于 $C$ 于 $A, B$ 两点, 则 $|A B|=$

Gaokao · Math · previous-year question

  1. A.$\frac{\sqrt{30}}{3}$
  2. B.6
  3. C.12correct
  4. D.$7 \sqrt{3}$

Answer

C. 12

Explanation

解: 由 $y^{2}=3 x$ 得其焦点 $F\left(\frac{3}{4}, 0\right)$, 准线方程为 $x=-\frac{3}{4}$. 则过抛物线 $y^{2}=3 x$ 的焦点 $F$ 且倾斜角为 $30^{\circ}$ 的直线方程为 $y=\tan 30^{\circ}\left(x-\frac{3}{4}\right)=\frac{\sqrt{3}}{3}$ $$ \left(x-\frac{3}{4}\right) $$ 代入抛物线方程, 消去 $y$, 得 $16 x^{2}-168 x+9=0$. 设 $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$ 则 $\mathrm{x}_{1}+\mathrm{x}_{2}=\frac{168}{16}=\frac{21}{2}$, 所以 $|A B|=x_{1}+\frac{3}{4}+x_{2}+\frac{3}{4}=\frac{3}{4}+\frac{3}{4}+\frac{21}{2}=12$ 故选: C.

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