题目内容
设F为抛物线y2=4x的焦点,A,B,C为该抛物线上三点,若
+
+
=
,则|
|+|
|+|
|=______.
| FA |
| FB |
| FC |
| 0 |
| FA |
| FB |
| FC |
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
∵
+
+
=
,
∴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
故答案为:6.
抛物线焦点坐标F(1,0),准线方程:x=-1
∵
| FA |
| FB |
| FC |
| 0 |
∴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
故答案为:6.
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