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[几何] 椭圆的密切圆

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hbghlyj Post time 2024-3-21 10:43 |Read mode
青青子衿 发表于 2020-7-30 10:38
椭圆的曲率圆:
\[ \left(x-\frac{a^{2}-b^{2}}{a}\left(\cos t\right)^{3}\right)^{2}+\left(y-\frac{a^{2}-b^{2}}{b}\left(\sin t\right)^{3}\right)^{2}=\frac{\left(\left(a\sin t\right)^{2}+\left(b\cos t\right)^{2}\right)^{3}}{a^{2}b^{2}} \]

上面 @青青子衿 给出的是椭圆$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$在$(x,y)=(a\cos(t),\color{red}-b\sin(t))$的密切圆。注意有个负号!

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 Author| hbghlyj Post time 2024-3-21 11:11
验证:
  1. eq1=(x-((a^2-b^2) Cos[t]^3)/a)^2-(b^2 Cos[t]^2+a^2 Sin[t]^2)^3/(a^2 b^2)+(y-((a^2-b^2) Sin[t]^3)/b)^2;
  2. eq2=x^2/a^2+y^2/b^2-1;
  3. Series[Resultant[eq1,eq2,y],{x,a Cos[t],2}]
Copy the Code

输出$O\left((x-a \cos (t))^3\right)$
  1. Series[Resultant[eq1,eq2,x],{y,-b Sin[t],2}]
Copy the Code

输出$O\left((y+b \cos (t))^3\right)$
交点是3阶的,所以确实是椭圆在$(a\cos(t),-b\sin(t))$的密切圆

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 Author| hbghlyj Post time 2024-3-21 11:31
hbghlyj 发表于 2024-3-21 02:43
椭圆$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$在$(x,y)=(a\cos(t),\color{red}-b\sin(t))$的密切圆。注意有个负号!


如果要把负号去掉,只需把$t$换成$-t$,得出:
$$\left(x-\frac{a^{2}-b^{2}}{a}\left(\cos t\right)^{3}\right)^{2}+\left(y\color{red}+\frac{a^{2}-b^{2}}{b}\left(\sin t\right)^{3}\right)^{2}=\frac{\left(\left(a\sin t\right)^{2}+\left(b\cos t\right)^{2}\right)^{3}}{a^{2}b^{2}}$$
是椭圆$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$在$(x,y)=(a\cos(t),b\sin(t))$的密切圆。

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 Author| hbghlyj Post time 2024-3-21 11:31
推导过程:
$$x=a\cos(t),y=b\sin(t)$$
$$\dot x=-a\sin(t),\dot y=b\cos(t)$$
$$\ddot x=-a\cos(t),\ddot y=-b\sin(t)$$
$$\kappa=\frac{\dot x\ddot y-\ddot x\dot y}{(\dot x^2+\dot y^2)^{3/2}}=\frac{ab}{(a^2\sin^2(t)+b^2\cos^2(t))^{3/2}}$$
$$r=\frac1\kappa=\frac{\left(\left(a\sin t\right)^{2}+\left(b\cos t\right)^{2}\right)^{3/2}}{ab}$$
法向量$=(-\dot y,\dot x)=(-b\cos(t),-a\sin(t))$
单位法向量$=\frac1{(a^2\sin^2(t)+b^2\cos^2(t))^{1/2}}(-b\cos(t),-a\sin(t))$
曲率中心$=(a\cos(t),b\sin(t))+r\cdot\text{单位法向量}=$
$$=(a\cos(t),b\sin(t))+\frac{\left(\left(a\sin t\right)^{2}+\left(b\cos t\right)^{2}\right)^{3/2}}{ab}\frac1{(a^2\sin^2(t)+b^2\cos^2(t))^{1/2}}(-b\cos(t),-a\sin(t))$$
$$=(a\cos(t),b\sin(t))+\frac{\left(a\sin t\right)^{2}+\left(b\cos t\right)^{2}}{ab}(-b\cos(t),-a\sin(t))$$
$$=\left(\frac{\left(a^2-b^2\right) \cos ^3(t)}{a},-\frac{\left(a^2-b^2\right) \sin ^3(t)}{b}\right)$$

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