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(1/2)已知倾斜角为X的直线过抛物线y^2=2px(x>0)的焦点F,与抛物线交于A.B二点.求证.|AB|=2p/sin^2x S三
人气:370 ℃ 时间:2020-03-27 18:49:33
解答
将直线y=tanα*(x-p/2)代入y^2=2px
(tanα)^2x^2-[(tanα)^2+2]px+(ptanα)^2/4=0
|AB|=|AF|+|BF|
=x1+x2+p
=2p[1+1/(tanα)]^2
=2p/(sinα)^2
当AP倾斜角为π/2时
|AB|=2p=2p/[sin(π/2)]^2
所以结论成立.
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