Question

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.

Answer 1

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