Question

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

Answer 1

不妨设p>q,抛物线准线方程L：x=-1/4a,做PE⊥L于E,QH⊥L于H,L交轴于F',直线PQ交L于G,FF'=1/2a,则

PE=PF=p

QH=QF=q

GP/PE=GQ/QH,即

GP/p=(GP+p+q)/q

GP=p(p+q)/(q-p)

GP/PE=GF/FF',即

[p(p+q)/(q-p)]/p=(p(p+q)/(q-p)]+p)/(1/2a)

(p+q)/(q-p)=2a(p^2+pq+pq-p^2)/(q-p)

(p+q)/pq=4a

1/p+1/q=4a