求高手推导抛物线焦点弦长公式求推导AB=2p/sin²a(a为过焦点直线的倾斜角)
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求高手推导抛物线焦点弦长公式求推导AB=2p/sin²a(a为过焦点直线的倾斜角)
求高手推导抛物线焦点弦长公式
求推导AB=2p/sin²a(a为过焦点直线的倾斜角)
求高手推导抛物线焦点弦长公式求推导AB=2p/sin²a(a为过焦点直线的倾斜角)
(1)当直线的斜率不存在时,即a=90°
xA=xB=p/2
∴ yA=p,yB=-p
∴ |AB|=2p=2p/sin²90°
(2)当直线斜率存在时,k=tana
直线方程是y=k(x-p/2)
代入抛物线方程y²=2px
则k²(x-p/2)²=2px
∴ k²x²-(k²p+2p)x+k²p²/4=0
利用韦达定理,则xA+xB=(k²p+2p)/k²
利用抛物线定义
|AB|=|AF|+|BF|=xA+p/2+xB+p/2=xA+xB+p
即 |AB|=(k²p+2p)/k²+p
=2p+2p/k²
=2p(1+1/k²)
=2p*(1+cos²a/sin²a)
=2p*(sin²a+cos²a)/sin²a
=2p/sin²a
综上,|AB|=2p/sin²a
哈,我恰好会推导! 抛物线的焦点为F(p/2,0) 设直线l的方程为y=(x-p/2)tana (a≠90°),代入y²=2px 得y²tana-2py-p²tana=0 设A,B的坐标为(x1,y1)(x2,y2) y1+y2=2p/tana y1+y2=-p² 过A作x轴的垂线交过B与x轴平行的直线于点C AB=AC/sina=(y1-y2)/sina=1/sina √[(y1+y2)²-4y1y2]=1/sina √(4p²/tan²a+4p²)=2p/sin²a 当a=90°时,直线AB⊥Ox轴,这时sina=1,A、B的坐标分别为(p/2,p)(p/2,-p),AB=2p AB=2p/sin²a也成立