椭圆x^2/4+y^2=1.x=my+1相交与A.B两点.在X轴上存在点M.使得AM.BM斜率相乘为定值
2个回答
设A(my1+1,y1),B(my2+1,y2),M(p,0)
则：AM.BM斜率相乘=[y1/(my1+1-p)][(y2/(my2+1-p)]
=y1y2/[(my1+1-p)(my2+1-p)]
=y1y2/[m^2*y1y2+(1-p)m(y1+y2)+(1-p)^2]
=1/[m^2+(1-p)((y1+y2)/(y1y2))m+(1/(y1y2))(1-p)^2]
将x=my+1代入x^2/4+y^2=1,得：
(m^2+4)y^2+2my-3=0
y1,y2是此方程两根
y1+y2=-2m/(m^2+4)
y1y2=-3/(m^2+4)
所以,AM.BM斜率相乘=1/[m^2+(1-p)((y1+y2)/(y1y2))m+(1/(y1y2))(1-p)^2]
=1/[m^2+(1-p)(2m/3)m-((m^2+4)/3)(1-p)^2]
=-3/{m^2*[-3-2(1-p)+(1-p)^2]+4(1-p)^2}
=-3/{m^2*(p^2-4)+4(1-p)^2}
只要：p^2-4=0,则：AM.BM斜率相乘=-3/[4(1-p)^2]=定值
所以：p^2=4
p=2 或p=-2
所以,M坐标为：(2,0) 或(-2,0)
相关问题
