点差法专题

点差法专题例1:已知曲线与斜率2的直线相交于A,B两点,求弦AB中点的轨迹方程.12:C22yx),();,();,(:002211yxMyxByxA中点设解),(11yx),(22yx),(00yx则有:222222222121yxyx0))((2))((21212121yyyyxxxx022200ABKyx2KAB又04:00yx上式04:yxMABM例2:已知和椭圆外点P(0,2),过该点的直线与其相交于A,B两点,求弦AB中点的轨迹方程.12:C22yxAB),();,();,(:002211yxMyxByxA中点设解),(11yx),(22yx),(00yx则有:222222222121yxyx022200ABKyx0200xyKAB又0220000xyyx042:22yyxMM例3:已知椭圆过P(2,1)引一条弦AB恰好被该点平分,求此直线方程.1416:C22yx),();,();,(:002211yxPyxByxA中点设解16416422222121yxyx则有:024200ABKyx1;200yx又042ABK21ABK042:yxLABBAP),(22yx),(11yx),(00yx例4:已知椭圆中心在原点坐标轴为对称轴,它与直线x+y=1交于A,B两点,C是弦AB的中点且|AB|=;OC的斜率为求椭圆方程.2222AB),(11yx),(22yx),(00yxC解:设椭圆方程:mx2+ny2=1同时:),();,();,(002211yxCyxByxA中点设1122222121nymxnymx000ABKnymx1ABK又nmxynymx0000)1(..........2200OCKxynm同时:0)1(2)(11222nnxxnmnymxyx2214)2(2||2nmnnmnAB又)2(0322nmmnnm32;31)2();1(nm132322yx例:已知,求x+y2的最大值与最小值.14416922yx解:14416922yx3sinαy4cosαx设191622yx22sin9cos4yx9cos4cos9)cos1(9cos422)1,1(t;cost设949)(:2tttf原式985)(:92Maxtft时当4)(:1Mintft时当例:已知内一点,F是椭圆的左焦点,在椭圆上求一点M,使得|AM|+2|MF|最小,并求出最小值.1121622yx)3,2(A||||||2||MNAMMFAM时最小三点共线且与准线垂直当,,NMA212;4ecaNFMA解:1028||'AN)3,32(M'NM