抛物线的问题

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2007-11-11

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    解：设直线为y=x-p/2 代入抛物线方程：(x-p/2)²=2px x²-3px+p²/4=0 P1(x1,y1)、P2(x2,y2) x1+x2=3p x1x2=p²/4 由抛物线定义可知,|P1F|等于点P1到准线x=-p/2的距离d1, 而d1=x1+p/2。
     同理|P2F|等于点P2到准线x=-p/2的距离d2, d2=x2+p/2。
     所以1/P1F+1/P2F=(P1F+P2F)/P1FP2F ```````````````=(x1+x2+p)/P1FP2F ```````````````=4p/[(x1+p/2)(x2+p/2)] ```````````````=4p/[x1x2+(x1+x2)p/2+p²/4] ```````````````=4p/(p²/4+3p²/2+p²/4) ```````````````=2/p。

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