已知过抛物线的焦点F的直线与抛物线交于A,B两点,若A,B在抛物线的准线上的射影分别是A1,B1,则∠A1FB1=

90°

设抛物线方程为 y^2 = 2px
这样 AA1 = BB1 = p
AF = BF = p
A1F = B1F = √2p
A1B1 = 2
∴ A1F^2 + B1F^2 = A1B1^2
A1F⊥B1F

修改后 ：
设任意过焦点(p/2,0)的直线为 y = kx-p/2*k
代入y^2 = 2px
得
k^2*x^2 - (k^2*p + 2*p)*x + k^2*p^2/4 = 0;
设两交点坐标为(x1,y1)(x2,y2)

x1 + x2 = (k^2*p + 2*p)/k^2
x1*x2 = p^2/4

弦长L^2 = (x1-x2)^2 + (y1-y2)^2 = (k^2 + 1)*((x1+x2)^2 - 4x1x2) =
((p+2p/k^2)^2 - p^2]*(k^2 + 1)
L = √((p+2p/k^2)^2 - p^2]*(k^2 + 1)

设AB和A1B1成角度θ
则tanθ = k
sinθ = k/√(k^2 + 1);

A1B1 = L*sinθ = 2p/k*√(k^2 + 1)

A1F = √(y1^2 + (x1-p/2)^2)
B1F = √(y2^2 + (x2-p/2)^2)

A1F^2 + B1F^2 = (y1^2 + (x1-p/2)^2) + (y2^2 + (x2-p/2)^2
= (x1 + x2)^2 - 2x1x2 -p(x1+x2) + p^2/2 + k^2(x1+x2)^2 - 2(kx1-pk/2)(kx2-pk/2)

化简， 可看出A1F^2 + B1F^2 = L^2

A1F⊥B1F

解：如图：设准线与x轴的交点为K，

∵A、B在抛物线的准线上的射影为A1、B1，

由抛物线的定义可得，AA1=AF，

∴∠AA1F=∠AF