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[几何] 一道双曲线的离心率倒数问题

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力工 posted 2018-11-20 17:02 |Read mode
已知双曲线$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$的左焦点为$F$,取$x$轴上且位于双曲线右顶点右侧的一点$E$,以$FE$为直径作圆交双曲线的左右两支分别于点$M,N$,则$\dfrac{|FN|-|FM|}{|EF|}$的值为$\dfrac{a}{c}$.
这题显然可以用特殊点来处理。我算的时候用了用第二定义,并将圆与双曲线方程联立,得韦达表示式代入后求出的。
请教,有没有综合法就是几何法直接求?

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kuing posted 2018-11-20 17:44
见:forum.php?mod=viewthread&tid=5072 (还有椭圆的)

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original poster 力工 posted 2018-11-20 20:03
回复 2# kuing

感谢酷神!!

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