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[几何] 卡住了,一个椭圆问题

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realnumber posted 2013-11-11 22:36 |Read mode
已知动点P(x,y)在椭圆$\frac{x^2}{25}+\frac{y^2}{16}=1$上,点A(3,0),$\abs{\vv{AM}}=1$,且$\vv{PM}·\vv{AM}=0$,则$\abs{\vv{PM}}$的最小值为_____.
设三角坐标P(5cost,4sint),M(3+coss,sins)硬算?

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其妙 posted 2013-11-11 22:51
答案?
我都怀疑我自己了

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kuing posted 2013-11-11 22:55
只要 PA 最小就行了吧?这个可以算的

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original poster realnumber posted 2013-11-11 23:03
回复 3# kuing


    en,果然昏头了~~

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乌贼 posted 2013-11-11 23:04
Last edited by hbghlyj 2025-5-22 05:12$PM^2=AP^2-AM^2$,不知道$AP$在哪最小
211.png

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kuing posted 2013-11-11 23:08
而且这个题的数据特殊,A 刚好是焦点,所以居然是在右顶点处最小。
其实,只要 A 在对称轴上,都是可以算的。

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其妙 posted 2013-11-11 23:15
所以我都以为 我做错了呢

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