两定点坐标分别为A(-1,0),B(2,0),动点M满足∠MBA=2∠MAB,求动点M的轨迹方程.
1个回答

∠MBA=2∠MAB,设:M(x,y)

1、若∠MBA=90°,此时M(2,3)

2、若∠MBA≠90°,

设直线MA的斜率是k1=tan∠MAB=y/(x+1),直线MB的斜率是k2=tan(180°-∠MBA)=y/(x-2)

则:

∠MBA=2∠MAB

tan∠MBA=tan(2∠MAB)

tan∠MBA=[2tan∠MAB]/[1-tan²∠MAB]

-y/(x-2)=[2y/(x+1)]/[1-y²/(x+1)²]

化简,得:

(x+1)(2x-1)-y²=0 (y≠3)

综合(1)、(2),得:(x+1)(2x-1)-y²=0