Question

1、过抛物线y=ax2(a>0)的焦点F作一直线交抛物线交于P、Q两点，若线段PF、FQ的长分别为p 、q 则1/p+1/q=

最好是把解题过程也写出来，谢谢拉

Answer 1

x=y/a, ∴焦点F(0,1/4a),准线:y=-1/4a PQ斜率必然存在, 设PQ:y=kx+1/4a ∴y=ax=kx+1/4a ∴ax-kx-1/4a=0 由韦达定理 x1+x2=k/a, x1x2=-1/4a 由抛物线第二定义:抛物线上的点到焦点的距离的距离等于到准线的距离 ∴p=|PF|=P到准线距离=y1-(-1/4a)=y1+1/4a,q=|FQ|=Q到准线的距离=y2-(-1/4a)=y2+1/4a ∴1/p+1/q=1/(y1+1/4a)+1/(y2+1/4a)=1/(kx1+1/4a+1/4a)+1/(kx2+1/4a+1/4a) =1/(kx1+1/2a)+1/(kx2+1/2a)=[(kx2+1/2a)+(kx1+1/2a)]/(kx1+1/2a)(kx2+1/2a) =[k(x1+x2)+1/a]/[kx1x2+(k/2a)(x1+x2)+1/4a] =(k/a+1/a)/[-k/4a+k/2a+1/4a] =[(k+1)/a]/[(k+1)/4a] =4a