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original poster
lemondian
posted 2024-11-7 19:47
哦,是不是这样:
由链接里的$\frac{AB}{AM}=\frac{\cos(\alpha -\beta )}{\cos \alpha \cos \beta }=1+\tan \alpha \tan \beta $.
对于本题而言:
可得$\dfrac{2AO}{AM}=1-\lambda $.
从而可得$2\vv{AO}=(1-\lambda )\vv{AM}$。
则有$2(-x_A,-y_A)=(1-\lambda )(x_M-x_A,y_M-y_A)$.
解得$x_M=\dfrac{(1+\lambda )x_A}{\lambda -1},y_M=\dfrac{(1+\lambda )y_A}{\lambda -1}
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