过椭圆x²/16+y²/4=1 求（1）以p（2,-1）为中点的弦所在的直线方程

x^2/16+y^2/4=1
(1),弦AB
xA+xB=2xP=4,yA+yB=-2
k(AB)=(yA-yB)/(xA-xB)
[(xA)^2/16+(yA)^2/4]-[(xB)^2/16+(yB)^2/4]=1-1=0
(xA+xB)*(xA-xB)+4(yA+yB)*(yA-yB)=0
(xA+xB)+4(yA+yB)*(yA-yB)/(xA-xB)=0
k(AB)=(yA-yB)/(xA-xB)=-(xA+xB)/[4(yA+yB)]=-4/[4*(-2)]=0.5
下面的应该会了,如果不会就读文科
AB:
(2),弦中点E(x,y)
xA+xB=2xE=2x,yA+yB=2y
k(AB)=(yA-yB)/(xA-xB)=2
(3),弦中点D(x,y)
k(AB)=(yA-yB)/(xA-xB)=(yD-yQ)/(xD-xQ)=(y-2)/(x-8)