文档内容
母题突破 2 定点问题
母题 (2022·烟台模拟)已知椭圆C:+y2=1,点A(-2,0),直线l:y=kx+m与C交于P,
Q两点,且AP⊥AQ,证明:直线l过定点,并求出此定点的坐标.
思路分析
❶联立直线l与椭圆C方程
↓
❷求AP·AQ
↓
❸利用根与系数的关系化简AP·AQ=0,找到M与k的关系
↓
❹利用直线的点斜式方程求定点
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[子题1] 已知双曲线C:x2-=1(x>0),过右焦点F 的直线l 与曲线C交于A,B两点,设
2 1
直线l:x=,点D(-1,0),直线AD交l于M,求证:直线BM经过定点.
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[子题2] 已知椭圆C:+=1,过点(1,0)的两条弦PQ,MN相互垂直,若PQ=2PS,MN=
2MT,求证:直线ST过定点.
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规律方法 动线过定点问题的两大类型及解法
(1)动直线l过定点问题,解法:设动直线方程(斜率存在)为y=kx+t,由题设条件将t用k表
示为t=mk,得y=k(x+m),故动直线过定点(-m,0).
(2)动曲线C过定点问题,解法:引入参变量建立曲线 C的方程,再根据其对参变量恒成立,
令其系数等于零,得出定点.
1.(2022·开封模拟)已知抛物线C:y2=4x,S(t,4)为C上一点,直线l交C于M,N两点(与点S不重合),直线SM,SN分别与y轴交于A,B两点,且OA·OB=8,判断直线l是否恒过
定点?若是,求出该定点;若不是,请说明理由.
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2.(2022·德州质检)已知抛物线C:y2=4x的顶点是坐标原点O,过抛物线C的焦点作与x
轴不垂直的直线l交抛物线C于两点M,N,直线x=1分别交直线OM,ON于点A和点B,
求证:以AB为直径的圆经过x轴上的两个定点.
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