数学问题!~~

x^2=y/a
焦点(0,1/(4a)),准线y=-1/(4a)
直线y-1/4a=kx=ax^2-1/4a
ax^2-kx-1/(4a)=0
x1+x2=k/a,x1*x2=-1/(4a^2)
y1+y2=k(x1+x2)-1/2a=k^2/a-1/2a=(2k^2-1)/a
y1*y2=(kx1+1/4a)(kx2+1/4a)
=k^2x1x2+k*1/4a*(x1+x2)+1/16a^2
=-k^2/4a^2+k^2/4a^2+1/16a^2
=1/16a^2

到焦点距离等于到准线距离
所以p=x1+1/(4a)
q=x2+1/(4a)
1/p+1/q=1/[y1+1/(4a)]+1/[y2+1/(4a)]
={[y1+1/(4a)]+[y2+1/(4a)]}/[y1+1/(4a)][y2+1/(4a)]
=[y1+y2+1/(2a)]/[y1*y2+(y1+y2)*1/(4a)+1/(16a^2)]
=[(2k^2-1)/a+1/2a]/[1/16a^2+(2k^2-1)/4a^2+1/16a^2]
=[(4k^2-2+1)/2a]/[(4k^2-2+1)/(8a)]
=4a

选C

抛物线焦点F（0,1/4a）抛物线准线方程：Y=-1/4a,抛物线上的任意点到焦点的距离与到准线的距离相等。因为是选择题，我们可以考虑特殊情况，PQ是经过焦点并且平行于X轴的直线，那么P、Q两点到准线的距离都等于1/2a
所以1/p+1/q=(p+q)/pq=4a