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[几何] 两条直线是共轭直径,则一条直线是另一条直线的无穷远点的极线。

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hbghlyj posted 2024-3-26 00:39 |Read mode
椭圆$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$的证明:直线$$l_1:(a\cos(\theta)t,b\sin(\theta)t),t\inR$$和另一条直线$$l_2:\left(a\cos(\theta+\frac\pi2)t,b\sin(\theta+\frac\pi2)t\right)=\left(-a\sin(\theta)t,b\cos(\theta)t\right),t\inR$$是共轭直径,则$l_1$是$l_2$的无穷远点$[-a\sin(\theta),b\cos(\theta),0]$的极线:
$$\frac{-a\sin(\theta)x}{a^2}+\frac{b\cos(\theta)y}{b^2}=0z\iff \frac{y}{b\sin(\theta)}=\frac{x}{a\cos(\theta)}$$

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