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青青子衿
发表于 2021-1-11 17:29
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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kuing
发表于 2020-9-13 18:16
