题目

函数y=0.5(x-1)^2+0.5,已知F（1,1）P.Q在函数图象上,PQ过F点,求证：1/PF+1/QF=2

提问时间：2021-01-14

答案

设直线PQ是y-1=k(x-1),设P(x1,y1) Q(x2,y2)
联立：x^2-(2+2k)x+2k=0
x1+x2=2+2k x1x2=2k
|PF|=√(x1-1)^2+(y1-1)^2 由y1=1/2(x1-1)^2+1/2 (x1-1)^2=2y1-1
=√(2y1-1+(y1-1)^2)
=y1
同理：|QF|=y2
所以1/PF+1/QF
=1/y1+1/y2
=(y1+y2)/y1y2
=(k(x1-1)+1+k(x2-1)+1)/(k(x1-1)+1)(k(x2-1)+1)
=(k(x1+x2)+2-2k)/(k^2x1x2+(1-k)(x1+x2)+(1-k)^2)
=2(k^2+1)/(k^2+1)
=2
得证