Question

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,

Answer 1

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