Question

设点P是椭圆C:x^2/a^2+y^2/b^2=1（a>b>0）上的任意一点,过原点的直线L与椭圆相交于M、N两点

Answer 1

设L:y=mx,代入椭圆方程得b^2x^2+a^2m^2x^2=a^2b^2,

∴x^2=a^2b^2/(a^2m^2+b^2),

∴交点M,N的横坐标x=土ab/√(a^2m^2+b^2),纵坐标y=土abm/√(a^2m^2+b^2),

设P(x,y)是椭圆x^2/a^2+y^2/b^2=1上任意一点,则y^2=(a^2-x^2)b^2/a^2,

kPM*kPN=[y-abm/√(a^2m^2+b^2)][y+abm/√(a^2m^2+b^2)]/{[x-ab/√(a^2m^2+b^2)][x+ab/√(a^2m^2+b^2)]}

=[y^2-a^2b^2m^2/(a^2m^2+b^2)]/[x^2-a^2b^2/(a^2m^2+b^2)]

=[(a^2m^2+b^2)y^2-a^2b^2m^2]/[(a^2m^2+b^2)x^2-a^2b^2]

=[(a^2m^2+b^2)(a^2-x^2)b^2-a^4b^2m^2]/{a^2[(a^2m^2+b^2)x^2-a^2b^2]}

=b^2[a^2b^2-(a^2m^2+b^2)x^2]/{a^2[(a^2m^2+b^2)x^2-a^2b^2]}

=-b^2/a^2,

与点P及直线L无关.