专题六 第4讲 母题突破2 定点(定直线)问题--2024年高考数学复习二轮讲义
展开(1)求双曲线C的方程;
(2)过点P的动直线l与C的左、右两支分别交于A,B两点,若点M在线段AB上,满足eq \f(|AP|,|AM|)=eq \f(|BP|,|BM|),证明:点M在定直线上.
思路分析
❶利用离心率和eq \(PF1,\s\up6(—→))·eq \(PF2,\s\up6(—→))=6求方程
❷设直线方程y-1=kx-3并联立
❸利用比例关系eq \f(|AP|,|AM|)=eq \f(|BP|,|BM|)列式
❹将根与系数的关系代入化简
❺消去参数得点在定直线上
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[子题1] (2023·信阳模拟)已知椭圆C:eq \f(x2,4)+eq \f(y2,2)=1的左、右顶点分别为A1,A2,过点D(1,0)的直线l与椭圆C交于异于A1,A2的M,N两点.若直线A1M与直线A2N交于点P,证明:点P在定直线上,并求出该定直线的方程.
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[子题2] (2023·岳阳模拟)已知双曲线C:x2-eq \f(y2,3)=1,P为双曲线的右顶点,设直线l不经过P点且与C相交于A,B两点,若直线PA与直线PB的斜率之和为-1.证明:直线l恒过定点.
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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.(2023·襄阳模拟)过抛物线x2=2py(p>0)内部一点P(m,n)作任意两条直线AB,CD,如图所示,连接AC,BD并延长交于点Q,当P为焦点并且AB⊥CD时,四边形ACBD面积的最小值为32.
(1)求抛物线的方程;
(2)若点P(1,1),证明:点Q在定直线上运动,并求出定直线方程.
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2.(2023·全国乙卷)已知椭圆C:eq \f(y2,a2)+eq \f(x2,b2)=1(a>b>0)的离心率是eq \f(\r(5),3),点A(-2,0)在C上.
(1)求C的方程;
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(2)过点(-2,3)的直线交C于P,Q两点,直线AP,AQ与y轴的交点分别为M,N,证明:线段MN的中点为定点.
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