设F为抛物线y2=4X的焦点.A.B.C为该抛物线上三点,若FA+FB+FC=O.则∣FA∣+∣FB∣+∣FC∣=?

Question

设F为抛物线y2=4X的焦点.A.B.C为该抛物线上三点,若FA+FB+FC=O.则∣FA∣+∣FB∣+∣FC∣=?

Answer 1

设A(x1,y1),B(x2,y2),C(x3,y3)
抛物线焦点坐标F(1,0),准线方程：x=-1
∵FA+FB+FC=O
∴点F是△ABC重心
则x1+x2+x3=3
y1+y2+y3=0
而|FA|=x1-(-1)=x1+1
|FB|=x2-(-1)=x2+1
|FC|=x3-(-1)=x3+1
∴|FA|+|FB|+|FC|=x1+1+x2+1+x3+1=(x1+x2+x3)+3=3+3=6