已知双曲线 $C:\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\left( {a > 0,b > 0} \right)$ 的左、右焦点分别为 ${F_1}$，${F_2}$，离心率为 $3$，直线 $y = 2$ 与 $C$ 的两个交点间的距离为 $\sqrt 6 $．
【难度】
【出处】
2013年高考大纲卷（理）
【标注】
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求 $a$，$b$；标注答案$a = 1$，$ b= 2\sqrt 2$．解析本题考查双曲线的基本量．由题设知 $\dfrac{c}{a} = 3$，即\[\dfrac{{{a^2} + {b^2}}}{a^2} = 9,\]故 ${b^2} = 8{a^2}$．所以 $C$ 的方程为\[8{x^2} - {y^2} = 8{a^2}.\]将 $y = 2$ 代入上式，求得\[x = \pm \sqrt {{a^2} + \dfrac{1}{2}}.\]由题设知，$2\sqrt {{a^2} + \dfrac{1}{2}} = \sqrt 6 $，解得 ${a^2} = 1$．所以\[\begin{split}a &= 1,\\ b &= 2\sqrt 2 .\end{split}\]
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设过 ${F_2}$ 的直线 $l$ 与 $C$ 的左、右两支分别交于 $A$，$B$ 两点，且 $\left| {A{F_1}} \right| = \left| {B{F_1}} \right|$，证明：$\left| {A{F_2}} \right|$，$\left| {AB} \right|$，$\left| {B{F_2}} \right|$ 成等比数列．标注答案略解析根据题中条件，求出直线的斜率，进而求出所需长度即可．由（1）知，${F_1}\left( { - 3,0} \right)$，${F_2}\left( {3,0} \right)$，$C$ 的方程为\[8{x^2} - {y^2} = 8. \quad\cdots\cdots ① \]由题意可设 $l$ 的方程为 $y = k\left(x - 3\right),\left| k \right| < 2\sqrt 2$，代入 ① 并化简得\[\left( {{k^2} - 8} \right){x^2} - 6{k^2}x + 9{k^2} + 8 = 0.\]设 $A\left( {{x_1},{y_1}} \right)$，$B\left( {{x_2},{y_2}} \right)$，则 ${x_1} \leqslant - 1$，${x_2} \geqslant 1$，\[\begin{split}{x_1} + {x_2} &= \dfrac{{6{k^2}}}{{{k^2} - 8}}, \\ {x_1} \cdot {x_2} &= \dfrac{{9{k^2} + 8}}{{{k^2} - 8}}.\end{split}\]于是\[\begin{split}\left| {A{F_1}} \right| &= \sqrt {{{\left( {{x_1} + 3} \right)}^2} + {y_1}^2} \\&= \sqrt {{{\left( {{x_1} + 3} \right)}^2} + 8{x_1}^2 - 8} \\&= - \left(3{x_1} + 1\right), \\ \left| {B{F_1}} \right| &= \sqrt {{{\left( {{x_2} + 3} \right)}^2} + {y_2}^2} \\&= \sqrt {{{\left( {{x_2} + 3} \right)}^2} + 8{x_2}^2 - 8} \\&= 3{x_2} + 1.\end{split}\]由 $\left| {A{F_1}} \right| = \left| {B{F_1}} \right|$，得 $ - \left( {3{x_1} + 1} \right) = 3{x_2} + 1$，即\[{x_1} + {x_2} = - \dfrac{2}{3},\]故\[\dfrac{{6{k^2}}}{{{k^2} - 8}} = - \dfrac{2}{3},\]解得 ${k^2} = \dfrac{4}{5}$，从而 ${x_1} \cdot {x_2} = - \dfrac{19}{9}$．由于\[\begin{split}\left| {A{F_2}} \right| &= \sqrt {{{\left( {{x_1} - 3} \right)}^2} + {y_1}^2} \\&= \sqrt {{{\left( {{x_1} - 3} \right)}^2} + 8{x_1}^2 - 8} \\&= 1 - 3{x_1} ,\\ \left| {B{F_2}} \right| &= \sqrt {{{\left( {{x_2} - 3} \right)}^2} + {y_2}^2} \\&= \sqrt {{{\left( {{x_2} - 3} \right)}^2} + 8{x_2}^2 - 8} \\&= 3{x_2} - 1, \end{split}\]故\[\begin{split}\left| {AB} \right| = \left| {A{F_2}} \right| - \left| {B{F_2}} \right| = 2 - 3\left( {{x_1} + {x_2}} \right) = 4, \\ \left| {A{F_2}} \right| \cdot \left| {B{F_2}} \right| = 3\left( {{x_1} + {x_2}} \right) - 9{x_1}{x_2} - 1 = 16.\end{split}\]因而\[\left| {A{F_2}} \right| \cdot \left| {B{F_2}} \right| = {\left| {AB} \right|^2},\]所以 $\left| {A{F_2}} \right|$，$\left| {AB} \right|$，$\left| {B{F_2}} \right|$ 成等比数列．
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