设双曲线X^2/a^2-Y^2/b^2=1的顶点为A1(-a0),A2(a0)设双曲线X^2/a^2-Y^2/b^2=1的顶点为A1(-a0),A2(a0),弦PQ垂直于A1A2,求A1P与A2Q交点的轨迹.

Question

设双曲线X^2/a^2-Y^2/b^2=1的顶点为A1(-a 0),A2 (a 0)
设双曲线X^2/a^2-Y^2/b^2=1的顶点为A1(-a 0),A2 (a 0),弦PQ垂直于A1A2,求A1P与A2Q交点的轨迹.

Answer 1

设P(x0,y0),A1(-a,0),A2(a,0)
PA1斜率=y0/(x0+a)=1/2,PA2斜率=y0/(x0-a)=2
y0=x0/2+a/2,y0=2x0-2a,x0=5a/3,y0=4a/3
P点在双曲线上,[(5a/3)^2]/a^2-[(4a/3)^2]/b^2=1,a^2/b^2=1
y=x y=-x
打字不易,