题目

高中解几抛物线题
三角形ABC内接于抛物线y^2=16x,其中A（1,4）,且三角形ABC的重心为抛物线的焦点,求直线BC的方程.

提问时间：2021-10-05

答案

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