题目

过抛物线y^2=2px（p>0)的焦点F做倾斜角为α的直线与抛物线交于A,B两点,求证：｜AB｜=2p/(sinα)^2

提问时间：2020-10-16

答案

y^2=4x
焦点F(p/2,0)
准线x=-p/2
设焦点弦:y=tanα*(x-p/2) (α≠π/2)
y=tanα*(x-p/2)代入y^2=2px
(tanα)^2x^2-[(tanα)^2+2]px+(ptanα)^2/4=0
由根与系数关系
x1+x2=p[(tanα)^2+2]/(tanα)^2=[1+2/(tanα)^2]p
由抛物线上任意一点到焦点距离与到准线距离相等
|AB|=|AF|+|BF|
=|x1+p/2|+|x2+p/2|
=x1+x2+p
=2p[1+1/(tanα)]^2
=2p[1+(cotα)^2]
=2p(cscα)^2
=2p/(sinα)^2
当AP倾斜角为π/2时
|AB|=2p=2p/[sin(π/2)]^2
可知|AB|=2p/(sinα)^2