若椭圆b2x2 a2y2=a2b2的一弦中点与原点连线及弦所在直线的斜率均存在,求证:两斜率之积为-b2/a2.

弦AB中点M
k(OM)=yM/xM=[(yA+yB)/2]/[(xA+xB)/2]=(yA+yB)/(xA+xB)
k(AB)=(yA-yB)/(xA-xB)
b^2x^2+a^2y^2=a^2b^2
[b^2(xA)^2+a^2(yA)^2]-[b^2(xB)^2+a^2(yB)^2]=0
b^2*(xA+xB)*(xA-xB)+a^2*(yA+yB)*(yA-yB)=0
b^2+a^2*[(yA+yB)/(xA+xB)]*[(yA-yB)/(xA-xB)]=0
k(OM)*k(AB)=-b^2/a^2

椭圆b2x2 + a2y2 = a2b2设弦的两个端点分别为A(x1，y1)，B(x2，y2)，设弦AB的中点为M(x0，y0)，这样2x0 = x1 + x2和2y0 = y1 + y2，所以b2x12 + a2y12 = a2b2 ①，b2x22 + a2y22 = a2b2 ②，作差可得b2(x1 + x2) + a(y1 + y2)(y1 – y2)/(x1 – x2) = 0 => 2b...