﻿ 一类四次李氏卵圆

# 一类四次李氏卵圆A Kind of Quartic Li’s Oval

Abstract: In this paper, a definition of Li's oval is provided, and the definitions of the oval center, the long radius, the short radius, and the symmetrical radius of the oval are provided. A kind of quartic equation is given and proved to be an oval equation. This proves some decision theorems for some quartic equations to become the Li's oval equation, the tangent equation and normal equation of the oval and the volume formula of rotating oval ball is provided. Finally, an example is given to verify the effectiveness of the simulation.

1. 引言

1) 笛卡尔卵形线 [1] ：A，B是平面内两个定点，m，n，b是三个固定正数，平面内满足 $m\cdot PA+n\cdot PB=b$ 的点P的轨迹称为笛卡尔卵形线。若定义焦距2a，中点为原点，则笛卡尔卵形线方程为

$m\sqrt{{\left(x-a\right)}^{2}+{y}^{2}}+n\sqrt{{\left(x+a\right)}^{2}+{y}^{2}}=b$

2) 卡西尼卵形线 [2] ：A，B是平面内两个定点，a，c是两个正数， $AB=2c$ ，平面内满足 $PA\cdot PB={a}^{2}$ 的点P的轨迹称为卡西尼卵形线，即到两定点之距的乘积为常数的点的轨迹，其方程为

${\left({x}^{2}+{y}^{2}\right)}^{2}-2{c}^{2}\left({x}^{2}-{y}^{2}\right)={a}^{4}-{c}^{4}$

2. 定义

① L有一条对称轴且是闭合的；

② L处处光滑；

③ L上有唯一一对对称点，设为S，T到对称轴的距离最大；

④ L与其对称轴有且仅有两个交点，设为P，Q，又PQ与ST交于一点设为O，且 $OP>OQ$

⑤ L为凸曲线，且不含直线段。

Figure 1. Li’s Oval and its rectangular coordinate system

3. 一类四次李氏卵圆方程

$a,b,c>0$ ，且 $b (1)

$\frac{{x}^{2}}{{\left(\frac{a-b}{a+b}x+\frac{2ab}{a+b}\right)}^{2}}+\frac{{y}^{2}}{{c}^{2}}=1$ (2)

$y=±2c\sqrt{ab}\frac{\sqrt{-{x}^{2}+\left(a-b\right)x+ab}}{\left(a-b\right)x+2ab}$ (3)

${x}^{2}{y}^{2}+\frac{4ab}{a-b}x{y}^{2}+\frac{4ab{c}^{2}}{{\left(a-b\right)}^{2}}{x}^{2}+\frac{4{a}^{2}{b}^{2}}{{\left(a-b\right)}^{2}}{y}^{2}-\frac{4ab{c}^{2}}{a-b}x-\frac{4{a}^{2}{b}^{2}{c}^{2}}{{\left(a-b\right)}^{2}}=0$ (4)

${\left(a-b\right)}^{2}{x}^{2}{y}^{2}+4ab\left(a-b\right)x{y}^{2}+4ab{c}^{2}{x}^{2}+4{a}^{2}{b}^{2}{y}^{2}-4ab{c}^{2}\left(a-b\right)x-4{a}^{2}{b}^{2}{c}^{2}=0$ (5)

$\left\{\begin{array}{l}x=\frac{2ab\mathrm{cos}t}{a+b-\left(a-b\right)\mathrm{cos}t}\\ y=c\mathrm{sin}t\end{array},t\in \left[0,2\text{π}\right]$ (6)

$\frac{a-b}{a+b}x+\frac{2ab}{a+b}=\frac{a-b}{a+b}\cdot \frac{2ab\mathrm{cos}t}{a+b-\left(a-b\right)\mathrm{cos}t}+\frac{2ab}{a+b}$

$\frac{a-b}{a+b}x+\frac{2ab}{a+b}=\frac{2ab}{a+b-\left(a-b\right)\mathrm{cos}t}$ (7)

${x}^{2}={\left[\frac{2ab}{a+b-\left(a-b\right)\mathrm{cos}t}\right]}^{2}{\mathrm{cos}}^{2}t={\left(\frac{a-b}{a+b}x+\frac{2ab}{a+b}\right)}^{2}{\mathrm{cos}}^{2}t$

$\frac{{x}^{2}}{{\left(\frac{a-b}{a+b}x+\frac{2ab}{a+b}\right)}^{2}}={\mathrm{cos}}^{2}t$ (8)

$\frac{{y}^{2}}{{c}^{2}}={\mathrm{sin}}^{2}t$ (9)

(8)与(9)相加，并注意到 ${\mathrm{sin}}^{2}t+{\mathrm{cos}}^{2}t=1$ ，即得(2)。故(6)与(2)等价，从而(2)~(6)互相等价。

${{y}^{\prime }}_{x}\left(t\right)=-\frac{c}{2ab\left(a+b\right)}{\left[\left(a+b\right)-\left(a-b\right)\mathrm{cos}t\right]}^{2}\mathrm{cot}t$ (10)

${{y}^{″}}_{x}\left(t\right)=-\frac{c}{4{a}^{2}{b}^{2}{\left(a+b\right)}^{2}}\cdot h\left(t\right)\cdot g\left(t\right)$ (11)

$g\left(t\right)=-\frac{{\left[a+b-\left(a-b\right)\mathrm{cos}t\right]}^{3}}{{\mathrm{sin}}^{3}t},t\in \left[0,2\text{π}\right]$ (12)

$h\left(t\right)=a+b-3\left(a-b\right)\mathrm{cos}t+2\left(a-b\right){\mathrm{cos}}^{3}t,t\in \left[0,2\text{π}\right]$ (13)

${{y}^{\prime }}_{x}\left(t\right)=\frac{{{y}^{\prime }}_{t}\left(t\right)}{{{x}^{\prime }}_{t}\left(t\right)}=\frac{{\left(c\mathrm{sin}t\right)}^{\prime }}{{\left(\frac{2ab\mathrm{cos}t}{a+b-\left(a-b\right)\mathrm{cos}t}\right)}^{\prime }}$

${{y}^{″}}_{x}\left(t\right)=\frac{{\left[{{y}^{\prime }}_{x}\left(t\right)\right]}^{\prime }}{{{x}^{\prime }}_{t}}=\frac{{\left\{-\frac{c}{2ab\left(a+b\right)}{\left[a+b-\left(a-b\right)\mathrm{cos}t\right]}^{2}\mathrm{cot}t\right\}}^{\prime }}{{\left(\frac{2ab\mathrm{cos}t}{a+b-\left(a-b\right)\mathrm{cos}t}\right)}^{\prime }}$

${{y}^{″}}_{x}\left(t\right)=-\frac{c}{4{a}^{2}{b}^{2}{\left(a+b\right)}^{2}}\cdot \frac{{\left[a+b-\left(a-b\right)\mathrm{cos}t\right]}^{3}}{{\mathrm{sin}}^{3}t}\cdot \left[a+b-3\left(a-b\right)\mathrm{cos}t+2\left(a-b\right){\mathrm{cos}}^{3}t\right]$ (14)

$h\left(t\right)\ge 0,t\in \left[0,\text{π}\right]$ (15)

${h}^{\prime }\left(t\right)=3\left(a-b\right)\mathrm{sin}t\left(2{\mathrm{sin}}^{2}t-1\right)$ (16)

${h}^{″}\left(t\right)=3\left(a-b\right)\mathrm{cos}t\left(5-6{\mathrm{cos}}^{2}t\right)$ (17)

$t=\frac{\text{π}}{4}$ 代入(17)得

${h}^{″}\left(\frac{\text{π}}{4}\right)=3\sqrt{2}\left(a-b\right)>0$

$h\left(\frac{\text{π}}{4}\right)=a+b-\sqrt{2}\left(a-b\right)$

$h\left(t\right)\ge a+b-\sqrt{2}\left(a-b\right),t\in \left[0,\text{π}\right]$ (18)

$a\le \left(3+\sqrt{2}\right)b⇔a+b-\sqrt{2}\left(a-b\right)\ge 0$ (19)

$y\left(x\right)=2c\sqrt{ab}\frac{\sqrt{-{x}^{2}+\left(a-b\right)x+ab}}{\left(a-b\right)x+2ab}$ (20)

${y}^{\prime }\left(x\right)=-\frac{c{\left(a+b\right)}^{2}\sqrt{ab}x}{{\left[\left(a-b\right)x+2ab\right]}^{2}\sqrt{-{x}^{2}+\left(a-b\right)x+ab}}$ (21)

${{y}^{″}}_{x}\left(t\right)\le 0,t\in \left(0,\text{π}\right)$ (22)

$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{c}^{2}}=1$

$B,C,E>0$ ，且 $E\le BC$ (23)

${x}^{2}{y}^{2}+Bx{y}^{2}+C{x}^{2}+\frac{{B}^{2}}{4}{y}^{2}-Ex-\frac{BE}{4}=0$ (24)

$a=\frac{E+\sqrt{{E}^{2}+BCE}}{2C},b=\frac{-E+\sqrt{{E}^{2}+BCE}}{2C},c=\sqrt{\frac{E}{B}}$ (25)

${x}^{2}{y}^{2}+Bx{y}^{2}+C{x}^{2}+D{y}^{2}-Ex-F=0$ (26)

$B=\frac{4ab}{a-b},C=\frac{4ab{c}^{2}}{{\left(a-b\right)}^{2}},D=\frac{4{a}^{2}{b}^{2}}{{\left(a-b\right)}^{2}},E=\frac{4ab{c}^{2}}{a-b},F=\frac{4{a}^{2}{b}^{2}{c}^{2}}{{\left(a-b\right)}^{2}}$ (27)

$D=\frac{{B}^{2}}{4},F=\frac{BE}{4}$ (28)

$B=\frac{4ab}{a-b},C=\frac{4ab{c}^{2}}{{\left(a-b\right)}^{2}},E=\frac{4ab{c}^{2}}{a-b}$ (29)

$\begin{array}{l}a\le \left(3+2\sqrt{2}\right)b\\ ⇔E+\sqrt{{E}^{2}+BCE}\le \left(3+2\sqrt{2}\right)\left(-E+\sqrt{{E}^{2}+BCE}\right)⇔\left(4+2\sqrt{2}\right)E\le \left(2+2\sqrt{2}\right)\sqrt{{E}^{2}+BCE}\\ ⇔{\left(4+2\sqrt{2}\right)}^{2}{E}^{2}\le {\left(2+2\sqrt{2}\right)}^{2}\left({E}^{2}+BCE\right)⇔\left(24+16\sqrt{2}\right){E}^{2}\le \left(12+8\sqrt{2}\right)\left(E+BC\right)E\\ ⇔8\left(3+2\sqrt{2}\right)E\le 4\left(3+2\sqrt{2}\right)\left(E+BC\right)⇔2E\le E+BC⇔E\le BC\end{array}$

4. 四次李氏卵圆的切线方程和法线方程

$y-{y}_{0}=-\frac{2ab{c}^{2}{\left(a+b\right)}^{2}{x}_{0}}{{\left[\left(a-b\right){x}_{0}+ab\right]}^{3}{y}_{0}}\left(x-{x}_{0}\right)$ (30)

$y-{y}_{0}=\frac{{\left[\left(a-b\right){x}_{0}+2ab\right]}^{3}{y}_{0}}{2ab{c}^{2}{\left(a+b\right)}^{2}{x}_{0}}\left(x-{x}_{0}\right)$ (31)

$F\left(x,y\right)=\frac{{x}^{2}}{{\left(\frac{a-b}{a+b}x+\frac{2ab}{a+b}\right)}^{2}}+\frac{{y}^{2}}{{c}^{2}}-1=0$

${{y}^{\prime }}_{x}=-\frac{{F}_{x}}{{F}_{y}}=-\frac{2ab{c}^{2}{\left(a+b\right)}^{2}x}{{\left[\left(a-b\right)x+2ab\right]}^{3}y}$

$K\left({x}_{0},{y}_{0}\right)=-\frac{2ab{c}^{2}{\left(a+b\right)}^{2}{x}_{0}}{{\left[\left(a-b\right){x}_{0}+2ab\right]}^{3}{y}_{0}}$ (32)

5. 旋转卵球体的体积

$V=4\text{π}{c}^{2}\frac{ab}{a+b}\left[{\left(\frac{a+b}{a-b}\right)}^{3}\mathrm{ln}\frac{a}{b}-2\frac{a+b}{a-b}\right]$ (33)

$y=±c\sqrt{1-\frac{{x}^{2}}{{\left(\alpha x+\beta \right)}^{2}}}$ (34)

$\alpha =\frac{a-b}{a+b},\beta =\frac{2ab}{a+b}$ (35)

$V=2\text{π}{\int }_{-b}^{a}{y}^{2}\text{d}x$ (36)

$\frac{{x}^{2}}{{\left(\alpha x+\beta \right)}^{2}}=\frac{{x}^{2}}{{\alpha }^{2}{x}^{2}+2\alpha \beta x+{\beta }^{2}}=\frac{1}{{\alpha }^{2}}+\frac{{\beta }^{2}}{{\alpha }^{4}}\cdot \frac{1}{{\left(x+\frac{\beta }{\alpha }\right)}^{2}}-\frac{2\beta }{{\alpha }^{3}}\cdot \frac{1}{x+\frac{\beta }{\alpha }}$ (37)

$a=\frac{\beta }{1-\alpha },b=\frac{-\beta }{1+\alpha }$ (38)

$\begin{array}{c}V=\text{π}{c}^{2}{\int }_{\frac{-\beta }{1+\alpha }}^{\frac{\beta }{1-\alpha }}\left[1-\frac{1}{{\alpha }^{2}}-\frac{{\beta }^{2}}{{\alpha }^{4}}\cdot \frac{1}{{\left(x+\frac{\beta }{\alpha }\right)}^{2}}+\frac{2\beta }{{\alpha }^{3}}\cdot \frac{1}{x+\frac{\beta }{\alpha }}\right]\text{d}x\\ =\text{π}{c}^{2}{\left[\left(1-\frac{1}{{\alpha }^{2}}\right)x-\frac{{\beta }^{2}}{{\alpha }^{4}}\frac{1}{x+\frac{\beta }{\alpha }}+\frac{2\beta }{{\alpha }^{3}}\mathrm{ln}\left(x+\frac{\beta }{\alpha }\right)\right]}_{-\beta /\left(1+\alpha \right)}^{\beta /\left(1-\alpha \right)}\\ =\text{π}{c}^{2}\left[-\frac{2\beta }{{\alpha }^{2}}--\frac{2\beta }{{\alpha }^{2}}+-\frac{2\beta }{{\alpha }^{3}}\mathrm{ln}\left(\frac{1+\alpha }{1-\alpha }\right)\right]\end{array}$

$V=2\text{π}{c}^{2}\beta \left[-\frac{2}{{\alpha }^{2}}+\frac{2}{{\alpha }^{3}}\mathrm{ln}\left(\frac{1+\alpha }{1-\alpha }\right)\right]$ (39)

$\underset{b\to a}{\mathrm{lim}}V=2\text{π}{c}^{2}a\underset{\alpha \to 0}{\mathrm{lim}}\frac{\frac{1}{1+\alpha }+\frac{1}{1-\alpha }-2}{3{\alpha }^{2}}=2\text{π}{c}^{2}a\underset{\alpha \to 0}{\mathrm{lim}}\frac{2}{3\left(1-{\alpha }^{2}\right)}=\frac{4}{3}\text{π}a{c}^{2}$

6. 示例与仿真验证

6.1 示例

$\frac{{x}^{2}}{{\left(\frac{1}{9}x+\frac{40}{9}\right)}^{2}}+\frac{{y}^{2}}{9}=1$ (40)

Figure 2. Equation (40) graphics

6.2 仿真验证

(a) (b)

Figure 3. Drawing program of simulating

$y=±147\sqrt{37080}\frac{\sqrt{-{x}^{2}+26x+37080}}{26x+74160}$ (41)

7. 结语

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[2] 数学手册编写组. 数学手册[M]. 北京: 人民教育出版社, 1977: 392-393.

[3] 管贤根, 管杰. 多焦点圆及其椭圆和卵圆[J]. 图学学报, 2013, 34(2): 52-64.

[4] 宋业存, 祝燕琴. 一种生成卵形曲线的方法[J]. 工程图学学报, 2006, 27(1): 160-163.

[5] 张家平, 陈化新. 运用切线基线法测算卵形曲线要点解析[J]. 测绘工程, 2010, 19(1): 17-20.

[6] 刘绍学. 数学必修2 [M]. 北京: 人民教育出版社, 2007: 118-122.

[7] 刘绍学. 数学选修2-1 [M]. 北京: 人民教育出版社, 2007: 38-40.

[8] 同济大学应用数学系. 高等数学上册[M]. 第五版. 北京: 高等教育出版社, 2002: 155, 153, 149, 273, 133.

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