设直线x-y=4a与y^2=4ax交于两点A,B(a为定值）,C为抛物线上任意一点,求△ABC的重心的轨迹方程

Question

设直线x-y=4a与y^2=4ax交于两点A,B(a为定值）,C为抛物线上任意一点,求△ABC的重心的轨迹方程

Answer 1

x = y + 4a,
y² = 4a(y + 4a)
y² - 4ay - 16a² = 0
y₁ + y₂ = 4a
x₁ + x₂ = y₁ + 4a + y₂ + 4a = 12a
AB中点M(m,n),m = (x₁ + x₂)/2 = 6a,n =(y₁ + y₂)/2 = 2a
M(6a,2a)
设重心N(x,y),C(c²/(4a),c)
根据重心的性质,|CN| = 2|NM|
x - c²/(4a) = 2(6a - x) (i)
y - c = 2(2a - y) (ii)
从(i)(ii)中消去c,(3y - 4a)² = 12a(x + 4a)
也是抛物线