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[几何] 2016年北京卷文科19题椭圆

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2009 posted 2017-11-8 11:02 |Read mode
Last edited by hbghlyj 2025-5-4 08:59已知椭圆$C:x^2/a^2+y^2/b^2=1$过点A(2,0),B(0,1)两点.
(I)求椭圆C的方程及离心率;
(II)设P为第三象限内一点且在椭圆C上,直线PA与y轴交于点M,直线PB与x轴交于点N,求证:四边形ABNM的面积为定值.

问:第二问有没有几何解法?

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kuing posted 2017-11-8 12:06
Last edited by hbghlyj 2025-5-4 09:05显然伸缩变换不改变结论,于是我们可以把椭圆变成圆,并不妨设其半径为 $1$,如下图。

四边形 $ABNM$ 的面积为定值等价于 $AN\cdot BM$ 为定值,令 $\angle PON=\alpha$, $\angle POM=\beta$,则
\begin{align*}
OM&=\tan\angle OAM=\tan\frac\alpha2,\\
ON&=\tan\angle OBN=\tan\frac\beta2,
\end{align*}
于是
\[AN\cdot BM=(1+OM)(1+ON)=1+\tan\frac\alpha2+\tan\frac\beta2+\tan\frac\alpha2\tan\frac\beta2,\]
由 $\alpha+\beta=90\du$ 得
\[\tan\frac{\alpha+\beta}2=1 \iff \tan\frac\alpha2+\tan\frac\beta2=1-\tan\frac\alpha2\tan\frac\beta2,\]
所以 $AN\cdot BM=2$,即得证。

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游客 posted 2017-11-9 16:05
Last edited by 游客 2017-11-9 17:36设M,N点的坐标,直线用截距式方程,两直线交点在椭圆上,求出关系.(代数法)

好在是证明题,配方就比较明确.

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游客 posted 2017-11-10 10:50
Last edited by hbghlyj 2025-5-4 09:07
\[
\begin{aligned}
& \text {圆中, } \triangle NBA \sim \triangle ABP \Rightarrow \frac{NB}{AB}=\frac{AB}{PB} \\
& \Rightarrow NB \cdot PB=AB^2=AF^2 \Rightarrow \frac{NB}{PB}=\left(\frac{AF}{PB}\right)^2 \\
& \text {又: } \triangle BVP \sim \triangle AVF \Rightarrow \frac{S_{\triangle NBM}}{S_{\triangle PBI}}=\frac{NB}{PB}=\frac{AF^2}{PB^2}=\frac{S_{\triangle AMF}}{S_{\triangle PBM}} \\
& \Rightarrow S_{\triangle N B M}=S_{\triangle MIF} \Rightarrow S_{\text {四边形ABNM}}=S_{\triangle A B F}
\end{aligned}
\]

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kuing posted 2017-11-10 16:55
回复 4# 游客

这个不错,比我的更加几何,初中生都能接受

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kuing posted 2017-11-10 17:03
回复 4# 游客

照这图,也可以这样证:
\[\frac{\S{NBM}}{\S{ABM}}=\frac{ON}{OA}
=\frac{ON}{OB}=\tan\angle OBN=\tan\angle FAM
=\frac{\sin\angle FAM}{\sin\angle BAM}
=\frac{\S{AMF}}{\S{AMB}}
\riff \S{NBM}=\S{AMF}.\]

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