已知抛物线的顶点在坐标原点,焦点F在X轴的正半轴上,F为焦点,A,B,C为抛物线上的三点,且满足

根据题意,抛物线可表达为y² = 2px,p >0
F(p/2,0),准线x = -p/2
设A(a²/(2p),a),B(b²/(2p),b),C(c²/(2p),c)
按抛物线的定义,|FA| +|FB| + |FC| = A与准线的距离+B与准线的距离+C与准线的距离
= a²/(2p) - (-p/2) + b²/(2p) - (-p/2) + c²/(2p) - (-p/2)
= (a² + b² + c²)/(2p) + 3p/2 = 6 (1)
向量FA = (a²/(2p) - p/2,a)
向量FB = (b²/(2p) - p/2,b)
向量FC = (c²/(2p) - p/2,c)
三者之和为((a² + b² + c²)/(2p) - 3p/2,a + b +c)
(a² + b² + c²)/(2p) - 3p/2 = 0
(a² + b² + c²)/(2p) = 3p/2
代入(1):6p/2 = 3p = 6
p = 2
y² = 4x