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[几何] 求教一个有关椭圆的想法能否成立?

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lemondian posted 2017-12-18 15:21 |Read mode
过椭圆$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)$外一点P(m,n)作椭圆的两条切线,切点分别为$A(x_1,y_1),B(x_2,y_2)$,请问$x_1x_2+y_1y_2$是不是定值?如果是定值,则这个定值是多少?

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kuing posted 2017-12-18 15:37
显然不定啊,如果 P 非常接近椭圆,那么该值非常接近 OP^2,如果 P 非常远,那么该值必为负。

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original poster lemondian posted 2017-12-18 15:44
看得不懂(P非常远时)
抛物线有定值,我以为椭圆 ,双曲线也有。

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zhcosin posted 2017-12-19 16:53
显然不定啊,因为
\[ x_1x_2+y_1y_2 = \vec{OA} \cdot \vec{OB} \]
你要说这个可能是定值么?然而它可以是正值、负值、零值,看两个向量夹角情况了。
椭圆上随便找两个点$A$和$B$,只要它俩连线不过椭圆中心,则这两个点处的切线就必定相交于某个点$P$。

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original poster lemondian posted 2017-12-19 20:28
回复 4# zhcosin


    这个式好理解。
我所说的定值是指:结果只与a,b,m,n有关呀。

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色k posted 2017-12-19 20:46
…………

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zhcosin posted 2017-12-19 21:53
回复 6# 色k
哈哈,我也无语了

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