焦点为F的抛物线y2=4x有三点ABC△ABC的重心是F|FA||FB||FC|成等差数列则直线AC
人气:156 ℃ 时间:2020-01-27 14:43:31
解答
抛物线y^2=4x的焦点F为(1,0),
△ABC的重心是F,
∴xA+xB+xC=3xF=3,①
yA+yB+yC=3yF=0,
A,B,C在抛物线上,
∴|FA|=xA+1,|FB|=xB+1,|FC|=xC+1,
|FA||FB||FC|成等差数列,
<==>|FA|+|FC|=2|FB|,
<==>xA+xC=2xB,
代入①,xB=1.yB=土2,
∴xA+xC=2,yA+yC=干2.
设AC的方程是y=kx+m,代入y^2=4x,得
k^2x^2+(2km-4)x+m^2=0,
∴(4-2km)/k^2=2,
(4-2km)/k+2m=干2,
解得k=干2,m=土1,
∴AC:y=-2x+1,或y=2x-1.
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