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[几何] 截距范围

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v6mm131 Posted 2019-12-2 17:36 |Read mode
Last edited by hbghlyj 2025-4-9 22:24椭圆 $E: \frac{x^2}{4}+\frac{y^2}{3}=1$。若存在过 $A(a, a)$ 的两直线 $M N,  P Q$ 交椭圆 $E$ 于 $M,  N,  P,  Q$ 四点(其中 $a \in(-2,0) \cup(0,2)$)使得两直线斜率和为一常数,且点 $B, C$ 分别是线段 $M N, P Q$ 的中点。则直线 $B C$ 的纵截距的取值范围为 $\boxed{\left(-\frac{3}{2}, 0\right) \cup\left(0, \frac{3}{2}\right)}$

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2025-6-6 19:03 GMT+8

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