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
2个回答

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

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