题目

过抛物线y^2=2px的焦点F的弦AB,|FA|分之一 加|FB|等于P分之2

提问时间：2021-01-03

答案

设A(a,b),B(c,d)
因为过抛物线y^2=2px的焦点F的弦AB
所以
b^2=2pa.(1)
d^2=2pc.(2)
即
b^2/2p=a.(3)
d^2/2p=c.(4)
因为过抛物线y^2=2px的焦点F的弦AB
所以AF斜率=BF斜率 (点F(p/2,0))
b/(a-p/2)=d/(c-p/2)
bc-pb/2=ad-pd/2
bc-pb/2-ad+pd/2=0
bc-ad-p(b-d)/2=0
把（3）（4）代入得
b(d^2)/2p-(b^2)d/2p-p(b-d)/2=0
-bd(b-d)/2p-p(b-d)/2=0
bd(b-d)/2p+p(b-d)/2=0
(b-d)(bd/2p+p/2)=0
因为b≠d
所以bd/2p+p/2=0
bd=-(p^2).(5)
(3)×(4)得
ac=((bd)^2)/(4(p^2))
把（5）代入得
ac=((-(p^2))^2)/(4(p^2))
=(p^4)/(4(p^2))
=p^2/4.(6)
因为过抛物线y^2=2px的焦点F的弦AB
由抛物线第二定义得
|FA|=|A到x=-p/2的距离|=a+p/2
|FB|=|B到x=-p/2的距离|=c+p/2
所以1/|FA|+1/|FB|=1/(a+p/2)+1/(c+p/2)
=(a+c+p)/(ac+(a+c)p/2+p^2/4)
把(6)代入得
=(a+c+p)/(p^2/4+(a+c)p/2+p^2/4)
=(a+c+p)/(p^2/2+(a+c)p/2)
=2/p