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[几何] 来自减压群v6的一道椭圆

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kuing Posted at 2018-1-27 21:12:58 |Read mode
v6mm131(4744*****) 19:52:07
QQ图片20180127211132.png
大家试试这个题的搞法

QQ截图20180127211152.png
解:沿坐标轴方向作“伸缩变换”将椭圆变成圆,如图,则所有直线的斜率都变为原来的一半,那么变换后依然有 $k_{AD}:k_{CB}=2:1$,作 $EH\perp AD$ 于 $H$,则有
\begin{align*}
k_{AD}:k_{CB}=2:1&\iff\tan\angle DAB=2\tan\angle CBA\\
&\iff \tan\angle DAB=2\tan\angle CDA\\
&\iff \frac{EH}{HA}=2\cdot\frac{EH}{HD}\\
&\iff HD=2HA\\
&\iff EB=2EA,
\end{align*}
故此,对于变换前,同样有 $EB=2EA$,易得 $E(-1/3,0)$,所以 $k=3$。

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shidilin Posted at 2018-1-27 22:11:43
话说“减压群”是个什么群啊?

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singular Posted at 2018-1-27 22:49:17
常规思路也能做出来,需要转换一下

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 Author| kuing Posted at 2018-1-27 23:16:16
回复 2# shidilin

就是用来减压的[阴险]你懂的

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力工 Posted at 2018-1-28 09:57:48
回复 4# kuing

做题减压嘛,遇到难题有k神在,可以减少焦虑感。

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游客 Posted at 2018-1-28 11:05:47
用直线AD、BC的方程,然后把C、D两点的坐标表示出来,最后D、F、C 三点共线,计算应该容易的吧?

未命名.PNG

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