Question

设直线l与抛物线y2=8x交于A,B两点,线段AB的中点为(1,-1),求直线l的方程

Answer 1

L与抛物线y^2=8x交于A,B两点,线段AB的中点M(1,-1)

xA+xB=2xM=2,yA+yB=2yM=-2

k(L)=(yA-yB)/(xA-xB)

y^2=8x

(yA)^2=8xA.(1)

(yB)^2=8xB.(2)

(1)-(2):

(yA+yB)*(yA-yB)=8(xA-xB)

(yA+yB)*(yA-yB)/(xA-xB)=8

-2k(L)=8

k(L)=-4

L:y+1=-4(x-1)

直线l的方程:y=-4x+3