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[几何] 2014湖北九题

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康锡 posted 2014-6-10 15:57 |Read mode
设椭圆和双曲线方程分别为 $x^2/a_1^2+y^2/b_1^2=1$, $x^2/a_2^2+y^2/b_2^2=1$,它们的离心率分别为 $e_1$, $e_2$,设 $PF_1=m$, $PF_2=n$,则依题意,由余弦定理有
\[(2c)^2=m^2+n^2-mn=\frac{(m+n)^2}4+\frac{3(m-n)^2}4=a_1^2+3a_2^2,\]
所以
\[\frac1{e_1^2}+\frac3{e_2^2}=4,\]
由柯西不等式,即得
\[\left(\frac1{e_1}+\frac1{e_2}\right)^2\leqslant \left(1+\frac13\right)\left(\frac1{e_1^2}+\frac3{e_2^2}\right)=\frac{16}3,\]
所以
\[\frac1{e_1}+\frac1{e_2}\leqslant \frac4{\sqrt3},\]
取等略。

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其妙 posted 2014-6-10 16:46
回复 1# 康锡
设椭圆和双曲线方程分别为$\dfrac{x^2}{a_1^2}+\dfrac{y^2}{b_1^2}=1$, $\dfrac{x^2}{a_2^2}+\dfrac{y^2}{b_2^2}=1$,

它们的离心率分别为 $e_1$, $e_2$,设 $PF_1=m$, $PF_2=n$,

则依题意,由余弦定理有:
$(2c)^2=m^2+n^2-mn=\dfrac{(m+n)^2}4+\dfrac{3(m-n)^2}4=a_1^2+3a_2^2,$

所以,$\dfrac1{e_1^2}+\dfrac3{e_2^2}=4,$

由柯西不等式,得$\left(\dfrac1{e_1}+\dfrac1{e_2}\right)^2\leqslant \left(1+\dfrac13\right)\left(\dfrac1{e_1^2}+\dfrac3{e_2^2}\right)=\dfrac{16}3,$

所以,$\dfrac1{e_1}+\dfrac1{e_2}\leqslant \dfrac4{\sqrt3}$,取等略。

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