过抛物线y=ax^2(a>0)的焦点F作一直线抛物线于P.Q两点,若线段PF与FQ的长分别为p,q,则(1/p)+(1/q)= 4a

题目

过抛物线y=ax^2(a>0)的焦点F作一直线抛物线于P.Q两点,若线段PF与FQ的长分别为p,q,则(1/p)+(1/q)= 4a

答案

x^2=1/a*y 焦点F(0,1/4a) 准线y=-1/4a
作一直线y=kx+1/4a
于P(x1,y1).Q(x2,y2)两点,
线段PF与FQ的长=到准线的距离
p=y1+1/4a q=y2+1/4a
1/p+1/q=(y1+y2+1/2a)/(y1y2+1/4a(y1+y2)+1/16a^2)
[(y-1/4a)/k]^2=1/a*y整理得
y^2-(1/2a+1/k^2a)y+1/16a^2=0
y1+y2=1/2a+1/k^2a
y1y2=1/16a^2 代入
1/p+1/q=(y1+y2+1/2a)/(y1y2+1/4a(y1+y2)+1/16a^2)
=(1/a+1/k^2a)/(1/8a^2+1/8a^2+1/4k^2a^2)
=(1/a+1/k^2a)/(1/4a^2+1/4k^2a^2)
=(1/a+1/k^2a)/[(1/a+1/k^2a)*1/4a]
=4a