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[几何] 圆锥曲线 弧长 Graves定理

本帖最后由 hbghlyj 于 2021-2-3 10:47 编辑

QQ图片20200913171428.jpg
2020-9-13 17:15

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2020-9-13 17:36

抛物线情形的证明:
点A$(x_0,y_0)$在$y^2=2kx+k^2$(因为点A在抛物线外,所以k>1),过A作$y^2=2x+1$的两条切线,切点纵坐标满足方程
$y^2-2 y y_0+1+2 x_0=0$
故$y_B+y_C=2y_0,y_By_C=1+2x_0,y_B-y_C=2\sqrt{y_0^2-2x_0-1}$
先算弧长:
$\int_{y_C}^{y_B}\sqrt{\left(\frac{dx}{dy}\right)^2+1}dy=\int_{y_C}^{y_B}\sqrt{y^2+1}dy=\frac12\left[y\sqrt{y^2+1}\right]_{y_C}^{y_B}+\frac12\left[\operatorname{arcsinh} y\right]_{y_C}^{y_B}$
$\frac{1}{2} y_B \sqrt{y_B^2+1}-\frac{1}{2} y_C \sqrt{y_C^2+1}=\sqrt{\frac{k-1}{k}} \left(k+2 y_0^2\right)=M$
由3#,$\operatorname{arcsinh} y_C-\operatorname{arcsinh} y_B$是定值
然后算AB+AC:
$AB^2AC^2=4 \left(2 x_0-y_0^2+1\right)^2 \left(x_0^2+y_0^2\right)$
$AB^2+AC^2=4 \left(y_0^2-x_0\right) \left(y_0^2-2x_0-1\right)$
故$AB+AC=2 \sqrt{\left(y_0^2-2x_0-1\right) \left(-x_0+y_0^2+\sqrt{x_0^2+y_0^2}\right)}=2 \sqrt{\frac{k-1}{k}} \left(k+y_0^2\right)=N$
现在,$N-M=\sqrt{\frac{k-1}{k}}$是与点A的选取无关的定值.

椭圆的情况之前在这里 http://kuing.orzweb.net/redirect ... =5569&pid=28087 就有过讨论……

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c,d是共焦点的抛物线,点A在c上运动,由A引d的切线,切点为B,C,求证$\operatorname{arcsinh} y_C-\operatorname{arcsinh} y_B$是定值
证明:$\operatorname{arcsinh} y_C-\operatorname{arcsinh} y_B=\ln\frac{y_1+\sqrt{y_1^2+1}}{y_2+\sqrt{y_2^2+1}}$,故只需证$S=\left(y_1+\sqrt{y_1^2+1}\right)^2 \left(y_2-\sqrt{y_2^2+1}\right)^2+\left(y_1-\sqrt{y_1^2+1}\right)^2 \left(y_2+\sqrt{y_2^2+1}\right)^2$是与A无关的定值.
$S=-8 \left(2 x_0+1\right) \left(2 \sqrt{x_0^2+y_0^2}+1\right)+8 \left(2 x_0+1\right)^2+16 y_0^2+2=2 \left(8 k^2-8 k+1\right)$.所以S是与A无关的定值.

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本帖最后由 青青子衿 于 2021-1-11 18:31 编辑
故$AB+AC=2 \sqrt{\left(y_0^2-2x_0-1\right) \left(-x_0+y_0^2+\sqrt{x_0^2+y_0^2}\right)}=2 \sqrt{\frac{k-1}{k}} \left(k+y_0^2\right)=N$
现在,$N-M=\sqrt{\frac{k-1}{k}}$是与点A的选取无关的定值.
hbghlyj 发表于 2020-9-13 18:04

楼主主楼帖子最后一行敲错了

已知抛物线$\,\varGamma_1$:$y^2=2x+1$,则与$\,\varGamma_1$同焦同向的抛物线$\,\varGamma_2$:$y^2=2px+p^2$,
设点$A(x_A,y_A)$在抛物线$\,\varGamma_2$上,(因为点$A$在抛物线外,所以$p>1$),
过$A$作$\varGamma_1:y^2=2x+1$的两条切线,切点纵坐标$y_B$与$y_C$满足二次方程
\[\eta^2-2 \eta y_A+2 x_A+1=0\]
故\begin{align*}
\left\{\begin{split}
y_B+y_C&=2y_A\\
y_By_C&=2x_A+1\\
y_B-y_C&=2\sqrt{{y_A}^2-2x_A-1}
\end{split}\right.
\end{align*}
计算抛物线弧长:
\begin{align*}
\displaystyle{\stackrel{{\mbox{$\Large{\frown}$}}}{BC}}=\int_{y_C}^{y_B}\sqrt{1+\left(\dfrac{\mathrm{d}x}{\mathrm{d}y}\right)^2}\mathrm{d}y
=\int_{y_C}^{y_B}\sqrt{1+y^2}\mathrm{d}y=\dfrac{1}{2}\Big[\,y\sqrt{1+y^2}\,\Big]\!\bigg|_{y_C}^{y_B}+\dfrac{1}{2}\Big[\operatorname{arcsinh} y\Big]\!\bigg|_{y_C}^{y_B}
\end{align*}

\begin{align*}
\dfrac{1}{2} y_B \sqrt{1+{y_B}^2}-\dfrac{1}{2} y_C \sqrt{1+{y_C}^2}&=\dfrac{1}{2}\sqrt{{y_B}^2(1+{y_B}^2)+{y_C}^2(1+{y_C}^2)-2{y_By_C}\sqrt{(1+{y_B}^2)(1+{y_C}^2)}\,}\\
&=\sqrt{\frac{p-1}{p}} \left(p+2{y_A}^2\right)=M
\end{align*}

\begin{align*}
\dfrac{1}{2}\operatorname{arsinh}(y_B)-\dfrac{1}{2}\operatorname{arsinh}(y_C)&=\dfrac{1}{2}\ln\left(\dfrac{y_B+\sqrt{1+y_B}}{y_C+\sqrt{1+y_C}\,}\right)\\
&=\dfrac{1}{2}\ln\left(\xi\right)\\
&=\dfrac{1}{4}\ln\left(\dfrac{t+\sqrt{t^2-4}}{2}\right)\\
&=\dfrac{1}{4}\ln\left[8p^{2}-8p+1+4\left(2p-1\right)\sqrt{p\left(p-1\right)}\,\right]\\
&=\dfrac{1}{4}\ln\left[\left(\sqrt{p}+\sqrt{p-1}\right)^4\right]\\
&=\operatorname{arsinh}\left(\sqrt{p-1}\,\right)
\end{align*}

\begin{align*}
t&=\xi^2+\dfrac{1}{\xi^2}\\
&=\left(\dfrac{y_B+\sqrt{1+y_B}}{y_C+\sqrt{1+y_C}\,}\right)^2+\left(\dfrac{y_C+\sqrt{1+y_C}}{y_B+\sqrt{1+y_B}\,}\right)^2\\
&=\,2(8p^2-8p+1)
\end{align*}

然后算$\left|AB\right|+\left|AC\right|$:
\begin{align*}
\left|AB\right|^2\left|AC\right|^2&=\left[(x_A-x_B)^2+(y_A-y_B)^2\right]\left[(x_A-x_C)^2+(y_A-y_C)^2\right]\\
&=4\left({y_A}^2-2x_A-1\right)^2\left({x_A}^2+{y_A}^2\right) \\

\left|AB\right|^2+\left|AC\right|^2&=(x_A-x_B)^2+(y_A-y_B)^2+(x_A-x_C)^2+(y_A-y_C)^2\\
&=4\left({y_A}^2-2x_A-1\right)\left({y_A}^2-x_A\right)
\end{align*}

\begin{align*}
\left|AB\right|+\left|AC\right|&=\sqrt{\left|AB\right|^2+\left|AC\right|^2+2\sqrt{\left|AB\right|^2\left|AC\right|^2}\,}\\
&=2 \sqrt{\left({y_A}^2-2x_A-1\right) \left({y_A}^2-x_A+\sqrt{{x_A}^2+{y_A}^2\,}\,\right)}\\
&=2 \sqrt{\frac{p-1}{p}} \left(p+{y_A}^2\right)=N
\end{align*}

现在,$N-M=\color{red}{\sqrt{p(p-1)\,}}\,\,$是与点A的选取无关的定值.

MonomialList[((x - s)^2 + (y - u)^2) ((x - t)^2 + (y - v)^2), {x, y}]

$\left|AB\right|+\left|AC\right|-{\stackrel{{\mbox{$\Large{\frown}$}}}{BC}}=\sqrt{p(p-1)\,}-\operatorname{arsinh}\left(\sqrt{p-1}\,\right)$

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本帖最后由 hbghlyj 于 2021-2-3 11:57 编辑

论文下载
第五页的定理3.2就是1楼的定理,即Graves定理.
其推论是定理3.1,即Chasels定理.以n=2为例,椭圆$x_0,x_1,x_2$共焦点,过$x_0$上一点D作其切线被$x_1$反射,再被$x_2$反射,再被$x_3$反射回到D,则对$x_0$上任何一点重复这个操作可以回到它自身,而且$\triangle ABC$的周长恒定.
由Poncelet小定理,每次反射都得到$x_0$的切线.
在定理3.1中令n趋于无穷,就得到定理3.2.
3.png
2021-2-3 10:52

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Two Proofs of Graves's Theorem.pdf (529.53 KB)

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