抛物线Y＾2=4X，p(1,2)A(x1,y1)B(x2,y2)在抛物线上，PA与PB的斜率存在且倾斜角互补时，

tan(a)=(2-y1)/(1-x1)
tan(b)=(2-y2)/(1-x2)
tan(a+b)=(tan(a)+tan(b))/(1-tan(a)*tan(b))
PA与PB的倾斜角互补
所以0=tan(a)+tan(b)
即(2-y1)/(1-x1)+(2-y2)/(1-x2)=0
可得：(2-y1)(1-x2)+(2-y2)(1-x1)
=(2-y1)(1-1/4*y2^2)+(2-y2)(1-1/4*y1^2)
=1/4*(y1-2)(y2-2)(4+y1+y2)
=0
所以y1=2或y2=2或y1+y2=-4
因为PA与PB的斜率存在，所以y1=2或y2=2都舍去。
所以y1+y2=-4

(y2-y1)/(x2-x1)
=4*(y2-y1)/(y2^2-y1^2)
=4/(y2+y1)
=-1

A(x1,y1)B(x2,y2)在抛物线上，必然满足解析式Y＾2=4X，

y1^2=4x1, x1=y1^2/4,
y2^2=4x2, x2=y2^2/4,

即：A(y1^2/4 ,y1) B(y2^2/4,y2)

kPA=(y1-2)/(x1-1)=(y1-2)/(y1^2/4-1)=4/(y1+2),
kPB=(y2-2)/(x2-1)=(y2-2)/(y2^2/4-1)=4/(y2+2),

PA与PB的斜率存在且倾斜角互补,说明kPA=-kPB,
即：4/(y1+2)=-4/(y2+2),
y1+2=-(y2+2),
y1+y2=-4;

kAB=(Y2-Y1)/(X2-X1)
=(y2-y1)/(y2^2/4-y1^2/4)
=4/(y1+y2)
=4/(-4)
=-1.