三峡名校联盟2023年秋季联考高2026届数学.考卷答案

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三峡名校联盟2023年秋季联考高2026届数学.考卷答案试卷答案

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分析先求出双曲线的方程，再求出B，C的坐标，即可得出结论．

解答解：由题意，$\left\{\begin{array}{l}{\frac{9}{{a}^{2}}-\frac{4}{{b}^{2}}=1}\\{\frac{{a}^{2}+{b}^{2}}{{a}^{2}}=5}\end{array}\right.$，∴a²=8，b²=32，
∴双曲线的方程为$\frac{{x}^{2}}{8}-\frac{{y}^{2}}{32}$=1，
设B（x₁，y₁），C（x₂，y₂），
设AB的方程为y-2=k（x+3），代入双曲线方程，可得（4-k²）x²-2k（3k+2）x-（3k+2）²-32=0，
∴-3+x₁=$\frac{6{k}^{2}+4k}{4-{k}^{2}}$，
∴x₁=$\frac{3{k}^{2}+4k+12}{4-{k}^{2}}$，y₁=$\frac{2{k}^{2}+24k+8}{4-{k}^{2}}$，
∴B（$\frac{3{k}^{2}+4k+12}{4-{k}^{2}}$，$\frac{2{k}^{2}+24k+8}{4-{k}^{2}}$），
同理C（$\frac{3{k}^{2}-4k+12}{4-{k}^{2}}$，$\frac{2{k}^{2}-24k+8}{4-{k}^{2}}$）．
∴k_BC=$\frac{48}{8}$-6．
故答案为：6．

点评本题考查了双曲线的标准方程及其性质、直线与双曲线相交问题转化为方程联立可得根与系数的关系、斜率计算公式，考查了推理能力与计算能力，属于难题．