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Last edited by hbghlyj 2023-3-31 14:00抛物线$y^2=4(x+1)$上的点$A_i(i=1,\cdots,n)$, 射线$OA_i$的倾斜角为$\frac{2i-1}nπ$. 求证 $\displaystyle\sum_{i=1}^n|OA_i|=n^2$
| $n=1$ | \begin{tikzpicture}[x=.5cm,y=.5cm]\draw[domain=-4:4]plot({(\x)^2/4-1},\x);\fill(-1,0)circle(2pt);\fill[red](0,0)circle(2pt);\end{tikzpicture} | $n=2$ | \begin{tikzpicture}[x=.5cm,y=.5cm]\draw[domain=-4:4]plot({(\x)^2/4-1},\x);\draw[only marks,mark=*]plot coordinates{(90:2)(270:2)};\fill[red](0,0)circle(2pt);\end{tikzpicture}
| | $n=3$ | \begin{tikzpicture}[x=.5cm,y=.5cm]\draw[domain=-4:4]plot({(\x)^2/4-1},\x);\draw[only marks,mark=*]plot coordinates{(60:4)(180:1)(300:4)};\fill[red](0,0)circle(2pt);\end{tikzpicture} | $n=4$ | \begin{tikzpicture}[x=.5cm,y=.5cm]\draw[domain=-5:5]plot({(\x)^2/4-1},\x);\draw[only marks,mark=*]plot coordinates{(45:6.82843)(135:1.17157)(225:1.17157)(315:6.82843)};\fill[red](0,0)circle(2pt);\end{tikzpicture} | | $n=5$ | \begin{tikzpicture}[x=.5cm,y=.5cm]\draw[domain=-6.4:6.4]plot[smooth]({(\x)^2/4-1},\x);\draw[only marks,mark=*]plot coordinates{(36:10.47214)(108:1.52786)(180:1) (252:1.52786)(324:10.47214)};\fill[red](0,0)circle(2pt);\end{tikzpicture} |
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