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)=

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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======以下答案可供参考======
供参考答案1:
抛物线标准方程：x^2=y/a
F(0, 1/(4a)), 设P(x1,y1) Q(x2,y2),
PQ平行于X轴时，方程为：y=1/(4a), p=q=1/(2a), 1/p+1/q=4a
PQ不平行于X轴时，设其方程为x=k(y-1/(4a))
代入抛物线方程得：y=ak^2*(y-1/(4a))^2
16ay=k^2(4ay-1)^2
16(aky)^2-(8ak^2+16a)y+k^2=0
y1+y2=1/(2a) + 1/(ak^2)
y1y2=1/(16a^2)
准线：y=-1/(4a)
PF=y1+1/(4a), FQ=y2+1/(4a)
p+q=y1+y2+1/(2a)=(k^2+1)/(ak^2)
pq=y1y2+(y1+y2)/(4a)+1/(4a)^2
=1/(16a^2) + 1/(8a^2) + 1/(4a^2k^2) + 1/(16a^2)
=(k^2+1)/(4a^2k^2)
1/q+1/p=(p+q)/(pq)=4a
综上可知，等于4a.