过抛物线y=ax^2(a>0)的焦点F作直线交抛物线与P、Q两点,若PF与FQ的长分别为p、q,则1/p+1/q为多少

显然焦点为：（1/4a,0)
准线为y=-1/4a
设直线PQ为y=k(x-1/4a),P(x1,y1),Q(x2,y2)
将直线代入抛物线方程消去x
a(y/k+1/4a)²-y=0
ay²/k²+y(1/2k-1)+1/16a=0
y1+y2=-(1/2k-1)/a,y1*y2=1/16a²
由于抛物线上的点到焦点的距离等于到准线的距离
则p=y1+1/4a,q=y2+1/4a
则1/p+1/q=(y1+y2+1/2a)/[y1*y2+(y1+y2)/4a+1/16a²]
=(1-1/2k+1/2)/a/[1/16a²-(1/2k-1)/4a²+1/16a²]
=(3-k)/2a/[(3-k)/8a²]=4a