共焦点的圆锥曲线正交

hbghlyj

Confocal conic sections
以0为中心的圆和通过0的直线正交.

证明

验证\[∇\left(\frac yx\right)·∇(x^2+y^2)=0\]即可:

1. In[]:= D[x^2+y^2,{{x,y}}].D[y/x,{{x,y}}]//Factor
2. Out[]= 0

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共焦点的椭圆和双曲线正交.

证明

共焦点的椭圆或双曲线可以写成
\[\frac{x^2}{a^2-\lambda }+\frac{y^2}{b^2-\lambda }=1\]
上式为关于$\lambda$的二次方程, 有两个根
\[\lambda=\frac{a^2+b^2-x^2-y^2\pm\sqrt{\left(-a^2-b^2+x^2+y^2\right)^2-4 \left(a^2 b^2-a^2 y^2-b^2 x^2\right)}}2\]
验证\begin{align*}∇&\left(\frac{a^2+b^2-x^2-y^2+\sqrt{\left(-a^2-b^2+x^2+y^2\right)^2-4 \left(a^2 b^2-a^2 y^2-b^2 x^2\right)}}2\right)\\&·∇\left(\frac{a^2+b^2-x^2-y^2-\sqrt{\left(-a^2-b^2+x^2+y^2\right)^2-4 \left(a^2 b^2-a^2 y^2-b^2 x^2\right)}}2\right)=0\end{align*}即可:

1. In[]:= D[1/2 (a^2+b^2-x^2-y^2-Sqrt[(-a^2-b^2+x^2+y^2)^2-4 (a^2 b^2-b^2 x^2-a^2 y^2)]),{{x,y}}].D[1/2 (a^2+b^2-x^2-y^2+Sqrt[(-a^2-b^2+x^2+y^2)^2-4 (a^2 b^2-b^2 x^2-a^2 y^2)]),{{x,y}}]//Factor
2. Out[]= 0

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共焦点的开口相反的抛物线正交.

证明

\[y^2 - 2 x p - p^2=0\]上式为关于$p$的二次方程, 有两个根\[p=-x\pm\sqrt{x^2+y^2}\]验证\[∇\left(-x+\sqrt{x^2+y^2}\right)·∇\left(-x-\sqrt{x^2+y^2}\right)=0\]即可:

1. In[]:= D[-x+Sqrt[x^2+y^2],{{x,y}}].D[-x-Sqrt[x^2+y^2],{{x,y}}]//Factor
2. Out[]= 0

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hbghlyj

$z\mapsto \exp z$把平行于坐标轴的直线映射到以0为中心的圆和通过0的直线.
在Konforme Abbildungen第10页
$$u+\mathrm{i} v=\mathrm{e}^{x+\mathrm{i} y}=\mathrm{e}^{x} \mathrm{e}^{\mathrm{i} y}=\mathrm{e}^{x}(\cos y+\mathrm{i} \sin y)$$
Trennung in Real- und Imaginärteil ergibt:\begin{array}{l}u=\mathrm{e}^{x} \cos y \\ v=\mathrm{e}^{x} \sin y\end{array}Für die Netzlinien erhalten wir aus den senkrechten Geraden $x=\text{const.}=c$:\begin{array}{l}u=\mathrm{e}^{c} \cos y \\ v=\mathrm{e}^{\mathrm{c}} \sin y\end{array}Dies ist ein Kreis um den Ursprung mit dem Radius $\mathrm e^c$. Für $c = 0$ ergibt sich der Einheitskreis. Für die waagrechten Geraden $y = \text{const.}= d$ ergibt sich$$\frac{v}{u}=\frac{\sin d}{\cos d}=\tan d$$also $v = u \tan d$ ; dies ist ein Strahl vom Ursprung aus mit dem Steigungswinkel $d$.
Maple-Programm:

1. with(plots):
2. conformal(z,z=-Pi-Pi*I..Pi+Pi*I,grid=[19,19], scaling=constrained);
3. conformal(exp(z),z=-Pi-Pi*I..Pi+Pi*I,grid=[19,19], scaling=constrained);

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共焦抛物线网可以被认为是复平面的右半部分平行于坐标轴的直线网在共形映射 $z\mapsto z^2$ 下的像。 The net of confocal parabolas can be considered as the image of a net of lines parallel to the coordinate axes and contained in the right half of the complex plane by the conformal map $w=z^2$.

在topic10.pdf第2页有一个很精致的图
在topic1.pdf第14,15页也有精致的图.
见Konforme Abbildungen第6-8页:
$$u+\mathrm{i} v=(x+\mathrm{i} y)^{2}=x^{2}+2 \mathrm{i} x y-y^{2}$$
Trennung in Real- und Imaginärteil ergibt:
\begin{array}{l}u=x^{2}-y^{2} \\ v=2 x y\end{array}Wir berechnen nun die Bilder der achsenparallelen Geraden der $z$-Ebene, also die Bilder des „Karo-Netzes“ in der $z$-Ebene. Für die senkrechten Geraden $x = \text{const.}= c$ erhalten wir\begin{array}{l}u=c^{2}-y^{2} \\ v=2 c y\end{array}Aus der zweiten Gleichung folgt $y=\frac{v}{2 c}$, und durch Einsetzen in die erste Gleichung ergibt sich:$$u=c^{2}-\frac{v^{2}}{4 c^{2}}$$Dies ist die Gleichung einer liegenden, nach links offenen Parabel.

hbghlyj

Visual proof that confocal ellipses and hyperbolas intersect orthogonally, because each has a "reflection property". 共焦点的椭圆和双曲线正交, 还可以使用光学性质证明:$$2\alpha+2\beta=\pi$$
共焦点的开口相反的抛物线正交, 也可以使用光学性质证明:$$2\alpha+2\beta=\pi$$	略

hbghlyj

关于常见的结论“经过两点的任何一个圆和这两点的任何一个阿氏圆正交”
相关: 圆幂之比为定值 共轴圆组
可以验证梯度的内积为0:
\[∇\left(\frac{x^2+(y-i)^2}{x^2+(y+i)^2}\right)·∇\left(\frac{(x-1)^2+y^2}{(x+1)^2+y^2}\right)=0\]

1. D[(x^2+(y+I)^2)/(x^2+(y-I)^2),{{x,y}}].D[((x-1)^2+y^2)/((x+1)^2+y^2),{{x,y}}]//Simplify

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hbghlyj

hbghlyj 发表于 2023-1-20 22:40
共焦抛物线网可以被认为是复平面的右半部分平行于坐标轴的直线网在共形映射$z\mapsto z^2$下的像。

从平行于坐标轴的直线到共焦点的椭圆和双曲线的保角映射
见Chapter 9. Elementary Entire Functions
见Schwarz–Christoffel mapping这帖
见Wolfram functions的条目Sin

Images of a square grid of size $[-\pi/2,\pi/2]\times[-\pi/2,\pi/2]$ under the (conformal) map $z\mapsto\sin z$. The two "focal points" are the points $\pm1$.

见Wolfram functions的条目Cos

Images of a square grid of size $[-\pi/2,\pi/2]\times[-\pi/2,\pi/2]$ under the (conformal) map $z\mapsto\cos z$.

见20200526_000.pdf
(b) (See Goursat, §22, Example 2.) Let us now consider the function on the entire plane given in complex notation by
$$
f(x+i y)=\cos (x+i y)=\cos x \cosh y-i \sin x \sinh y .
$$
This function is analytic everywhere, and will be conformal everywhere that its derivative is nonzero. (We pause for a moment to clarify a point which the author fumbled during lecture. The derivative of $\cos z$ is $-\sin z$, which means that $\cos z$ will be conformal at every point where $\sin z$ is nonzero. Now
$$
\sin (x+i y)=\sin x \cosh y+i \cos x \sinh y
$$
and for this to be zero, we see first of all that we must have $\sin x=0$(since $\cosh y \geq 1$ for all real $y$ ), and since this means that $\cos x \neq 0$, we must have $\sinh y=0$, or $y=0$. Thus the zeros of $\sin z$ over the complex plane are the same as those of $\sin x$ over the real line, i.e., $n \pi, n \in \mathbf{Z}$. Thus $f$ will be conformal at every point inside the strip $\{x+i y \mid 0<x<\pi, y>0\}$. Let us consider how $f$ maps straight lines within this strip. Let us consider first a horizontal line, say $y=y_0>0$. On such a line, $f$ is equal to
$$
f\left(x+i y_0\right)=\cos x \cosh y_0-i \sin x \sinh y_0,
$$
where $x \in(0, \pi)$. Now this is just another way of writing the parametric curve
$$
t \mapsto\left(\cosh y_0 \cos t,-\sinh y_0 \sin t\right), \quad t \in(0, \pi) .
$$
If we denote this curve by $(x(t), y(t))$ (where unfortunately here $x(t)$ and $y(t)$ are completely distinct from the real and imaginary parts of $z$), then we have
$$
\left(\frac{x(t)}{\cosh y_0}\right)^2+\left(\frac{y(t)}{\sinh y_0}\right)^2=1,
$$
i.e., the curve must lie on an ellipse with major axis $\cosh y_0$ along the horizontal axis and minor axis sinh $y_0$ along the vertical axis, and centred at the origin. Now since $y_0>0$, $\sinh y_0>0$, so $-\sinh y_0<0$ and $y(t)<0$ for all $t \in(0, \pi)$, while $x(t)$ takes on all values from $\cosh y_0$ to $-\cosh y_0$. Thus we obtain the lower half of this ellipse.
Now let us consider a vertical line, say $x=x_0 \in(0, \pi)$. Working as before, we see that on this line
$$
f\left(x_0+i y\right)=\cos x_0 \cosh y-i \sin x_0 \sinh y .
$$
If $x_0=\pi / 2$ then $\cos x_0=0$ and this is simply a parametrisation of the negative imaginary axis. Otherwise, we again write
$$
(x(t), y(t))=\left(\cos x_0 \cosh t,-i \sin x_0 \sinh t\right), \quad t \in(0, \pi)
$$
and note that (this follows from the basic identity $\cosh ^2 x-\sinh ^2 x=1$)
$$
\left(\frac{x(t)}{\cos x_0}\right)^2-\left(\frac{y(t)}{\sin x_0}\right)^2=1,
$$
which means that the curve lies on a hyperbola opening along the real axis with intercept $\pm \cos x_0$ and with asymptotes having slope $\pm \tan x_0$. Now we note that $y(t)<0$ for all $t$, while $x(t)>0$ for $t \in(0, \pi / 2)$ and $x(t)<0$ for $t \in(\pi / 2, \pi)$; thus in the first case we have the lower right-hand portion of the hyperbola, while in the second case we have the lower left-hand portion. See Fig. 2. Note especially how the blue and red curves on the right intersect at right angles, exactly like those on the left.

hbghlyj

$z\mapsto\sqrt z$把平行于坐标轴的直线映射到双曲线族$x^2-y^2=k$和双曲线族$xy=k$
见Wolfram functions的条目Sqrt

Images of a square grid of size $[-2,2]\times[-2,2]$ under the (conformal) map $z\mapsto\sqrt z$.