以抛物线焦点为重心的三角形有几个

y^2 = 2*8x,焦点F(4,0)

AF的方程:(y - 4)/(x - 1) = (0-4)/(4 -1) = -4/3

4x + 3y -16 = 0

BC的中点M在直线AM上,M(m,(16-4m)/3)

|AF| = √[(4-1)² + (0-4)²] = 5

|AF| :|FM| = 2:1

|FM| = 5/2

|FM|² = 25/4 = (m - 4)² + [(16-4m)/3 - 0]² = (25/9)*(m-4)²

(m-4)² = 9/4

m-4 = ±3/2

m = 11/2,M(11/2,-2)

或m = 5/2,M(5/2,2) (与A都在轴上方,舍去)

B(b²/16,b)

C(c²/16,c)

BC的中点M((b²+c²)/32,(b+c)/2)

(b²+c²)/32 = 11/2

(b+c)/2 = -2

c = -2 ± √21

b = 2 ± √21

二者可互换,这里取C(另一值为B):

C((11-√21)/2,-2 + √21)

CM(即BC的方程):(y + 2)/(x - 11/2) = (-2 + 2√21 +2)/[11-√21)/2 -11/2] = -4

4x + y -20 = 0