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[几何] 又一道椭圆题!

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青青子衿 Posted 2014-1-23 15:08 |Read mode
Last edited by 青青子衿 2014-1-23 15:22已知椭圆$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,一条过点$P(x_0,y_0)$的直线交椭圆于$A,B$,使得$m|AP|=|BP|$,求直线方程。
能用点差法吗?

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007 Posted 2014-1-23 15:22
点差法一般适用于中点相关的
这个用点差法好像不容易处理哦

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其妙 Posted 2014-1-23 17:02
一般用直线的参数方程吧,不是特殊数据,难算哦

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