题目内容
设F为抛物线y2=2px(p>0)的焦点,A、B、C为该抛物线上三点,当
+
+
=
且|FA|+|FB|+|FC|=3时,此抛物线的方程为( )
| FA |
| FB |
| FC |
| 0 |
| A.y2=2x | B.y2=4x | C.y2=6x | D.y2=8x |
设向量FA FB FC分别为(x1,y1)(x2,y2)(x3,y3) 则x1+x2+x3=0
|FA|+|FB|+|Fc|=3
XA=x1+
,同理XB=x2+
,XC=x3+
|FA|=x2+
+
=x2+p
∴x1+x2+x3+3p=3
∴p=1
∴抛物线方程为y2=2x
故选A
|FA|+|FB|+|Fc|=3
XA=x1+
| p |
| 2 |
| p |
| 2 |
| p |
| 2 |
|FA|=x2+
| p |
| 2 |
| p |
| 2 |
∴x1+x2+x3+3p=3
∴p=1
∴抛物线方程为y2=2x
故选A
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