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

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  • 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)