曲线族的包络线

先来看两个问题。

第一个问题：

第二个问题： \begin{align} \\ &\text{试计算椭圆}\frac{x^2}{\left( 2-a \right) ^2}+\frac{y^2}{a^2}=1, a\in \left( 0,1 \right) \\ &\text{随参数}a\text{变化时图形扫过的面积。} \\ \end{align}

第二个问题用Desmos演示一下：

介绍一下背景知识。

\begin{align} \\ &\color{red}{\mathbf{曲线族的包络线}} \\ &\text{在几何学}, \text{某}个\text{曲线族的包络线}\left( \text{Envelope} \right) , 是\text{跟该曲线族的每条线都有至少一点} \\ &\text{相切的一条曲线。}\left( \text{曲线族即一些曲线的无穷集}, \text{它们有一些特定的关系。} \right) \\ \\ &\color{orange}{\mathbf{Exercise}\,\,\mathbf{1}}\,\,\text{一}个\text{梯子靠在墙上}, \text{由于梯子下沿比较光滑}, \text{致使梯子下滑直至滑倒在} \\ &\text{地}, \text{在这}个\text{过程中}, \text{求：} \\ &\left( 1 \right) \,\text{梯子中点的轨迹}; \ \left( 2 \right) \,\text{梯子靠近下沿的三等分点的轨迹}; \ \left( 3 \right) \,\text{梯子的包络线}. \\ &\color{orange}{\mathbf{Solution}\,\,\mathbf{1}}\,\,为\text{了方便}, \text{我们将梯子长度定}为\text{单位}1, 模\text{型抽象}为\text{如下坐标系}: \\ \end{align}

\begin{align} \\ &\left( 1 \right) \,\,\text{显然}AB\text{中点到原点的距离}为\frac{1}{2}, \text{即中点轨迹}是\text{圆}x^2+y^2=\frac{1}{4}; \\ &\left( 2 \right) \,\,\text{由于}\left| OA \right|^2+\left| OB \right|^2=1, \text{设}A\left( \cos \theta ,0 \right) ,\,B\left( 0,\sin \theta \right) ,\,\theta \in \left[ 0,\frac{\pi}{2} \right] \\ &\text{则靠近下沿的三等分点坐标}为\left( \frac{2}{3}\cos \theta , \frac{1}{3}\sin \theta \right) , \text{轨迹}为\text{椭圆}: \frac{9}{4}x^2+9y^2=1; \\ &\left( 3 \right) \,\,\text{设}A\left( t,0 \right) , \text{则}B\left( 0,\sqrt{1-t^2} \right) , \text{线段}AB\text{方程}为 y+\frac{\sqrt{1-t^2}}{t}\left( x-t \right) =0. \\ &\text{记}f\left( x,y,t \right) =y+\frac{\sqrt{1-t^2}}{t}\left( x-t \right) , \text{令}\frac{\partial}{\partial t}f\left( x,y,t \right) =0, \text{即} \\ &-\frac{\frac{t^2}{\sqrt{1-t^2}}+\sqrt{1-t^2}}{t^2}\left( x-t \right) \!-\frac{\sqrt{1-t^2}}{t}=0,\,\text{解得}t=x^{\frac{1}{3}},\text{代入方程得到包络线}为 \\ &\text{星形线}: \ \ x^{\frac{2}{3}}+y^{\frac{2}{3}}=1. \\ \\ \\ &\color{red}{\mathbf{总结}}:\,\text{对于曲线族}f\left( x,y,t \right) \!=0\,\left( t是\text{参数},\,\text{随}着t\text{取不同值得到不同的曲线} \right) ,\,\text{当包络} \\ &\text{线存在时},\,\text{包络线由方程组} \\ &\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{cases} \,\, f\left( x,y,t \right) =0\\ \frac{\partial}{\partial t}f\left( x,y,t \right) =0\\ \end{cases}\,\, \\ &\text{消去参数}t\text{得到的方程}h\left( x,y \right) =0\text{给出。} \\ &\text{这}个\text{很好理解}, \text{包络线与曲线族}f\left( x,y,t \right) =0\text{的每条曲线都相切}, \text{但}是\text{曲线族的包络} \\ &\text{线}是\text{固定的}, \text{不随}t\text{的改变而改变}, \text{这就}是\frac{\partial}{\partial t}f\left( x,y,t \right) =0. \\ \end{align}

\begin{align} \\ &\color{red}{\mathbf{拓展}}: \text{曲线族由方程}f\left( x,y,t \right) =0\,\text{给出}, t是\text{参数},\,\text{随}着t\text{取不同值得到不同的曲线。} \\ &\color{blue}{\left[ \mathbf{直线系} \right]} \,\,\text{过直线}l_1:A_1x+B_1y+C_1=0\text{与}l_2:A_2x+B_2y+C_2=0\text{交点的直线系方程}: \\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda _1\left( A_1x+B_1y+C_1 \right) +\lambda _2\left( A_2x+B_2y+C_2 \right) =0 \left( \lambda _1,\,\lambda _2\in \mathbb{R} \right) \\ &\color{blue}{\left[ \mathbf{圆系} \right]} \,\,两\text{圆}C_1,\,C_2\text{交于}A,\,B两\text{点},\,\text{其中}C_k:\,x^2\!\!+y^2\!+\!D_kx+E_ky+F_k=0\,\left( k=1,2 \right) ,\, \\ &\text{过}A,\!\:B两\text{点的圆系方程}为 \\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda _1\left( x^2\!\!+y^2\!+\!D_1x+E_1y+F_1=0 \right) +\lambda _2\left( x^2\!\!+y^2\!+\!D_2x+E_2y+F_2=0 \right) =0 \\ &\left( \lambda _{1}^{2}+\lambda _{2}^{2}\ne 0\text{且}\lambda _1\ne -\lambda _2 \right) \\ &\text{当}\lambda _1=-\lambda _2\text{时}, \text{圆退化}为\text{直线}\left( \text{相交}弦 \right) \end{align}

\begin{align} \\ &\color{blue}{\left[ \mathbf{二}\mathbf{次曲线系} \right]} \\ &\color{orange}{\left[ \text{定理} \right]} \,\,\text{若}两\text{条二}次\text{曲线}C_1: F_1\left( x,y \right) =0, C_2: F_2\left( x,y \right) =0\text{交于不共线四点}, \text{则过这} \\ &\text{四}个\text{交点的二}次\text{曲线系}为 \\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda _1F_1\left( x,y \right) +\lambda _2F_2\left( x,y \right) =0 \left( \lambda _{1}^{2}+\lambda _{2}^{2}\ne 0 \right) \\ &\text{包含了}所\text{有过这四}个\text{交点的二}次\text{曲线}\left( \text{圆、椭圆、双曲线、抛物线} \right) \\ &\color{orange}{\left[ \text{推论}1 \right]} \,\,\text{若}两\text{直线}l_1\left( x,y \right) =0, l_2\left( x,y \right) =0\text{与一条二}次\text{曲线}F\left( x,y \right) =0\text{相交于不共} \\ &\text{线四点}, \text{则过这四}个\text{交点的二}次\text{曲线系}为 \\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda F\left( x,y \right) +\mu l_1\left( x,y \right) \cdot l_2\left( x,y \right) =0 \left( \lambda ^2+\mu ^2\ne 0 \right) \\ &\color{orange}{\left[ \text{推论}2 \right]} \,\,\text{若有不共线四点}P_1,\,P_2,\,P_3,\,P_4, \text{直线}P_1P_2,\,P_2P_3,\,P_3P_4,\,P_4P_1\text{的方程分别}为 \\ &l_k\left( x,y \right) =0\,\,\,\left( k=1,2,3,4 \right) , \text{则过这四}个\text{交点的二}次\text{曲线系}为 \\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda _1l_1\left( x,y \right) \cdot l_3\left( x,y \right) +\lambda _2l_2\left( x,y \right) \cdot l_4\left( x,y \right) =0 \left( \lambda _{1}^{2}+\lambda _{2}^{2}\ne 0 \right) \\ &\color{orange}{\left[ \text{推论}3 \right]} \,\,\text{若有不共线三点}P_1,\,P_2,\,P_3, \text{直线}P_1P_2,\,P_2P_3,\,P_3P_1\text{的方程分别}为l_k\left( x,y \right) =0 \\ &\left( k=1,2,3 \right) , \text{则过这三}个\text{交点的二}次\text{曲线系}为 \\ &\ \ \ \ \ \ \lambda _1l_1\left( x,y \right) \cdot l_2\left( x,y \right) +\lambda _2l_2\left( x,y \right) \cdot l_3\left( x,y \right) +\lambda _3l_3\left( x,y \right) \cdot l_1\left( x,y \right) =0 \left( \lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}\ne 0 \right) \\ \end{align}

现在我们来解决文章开始的两个问题。

先计算一组曲线族的包络线：

\begin{align} \\ &\color{orange}{\mathbf{Solution}\,\,\mathbf{2}}\,\,\text{设}A\left( t,0 \right) , \text{则}B\left( \frac{1}{2}\left( 1-t \right) , \frac{\sqrt{3}}{2}\left( 1-t \right) \right) , \text{直线}AB\text{方程}为 \\ &\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y+\frac{\sqrt{3}\left( 1-t \right)}{3t-1}\left( x-t \right) =0. \\ &\text{记}f\left( x,y,t \right) =y+\frac{\sqrt{3}\left( 1-t \right)}{3t-1}\left( x-t \right) , \text{令}\frac{\partial}{\partial t}f\left( x,y,t \right) =0, \text{得到} \\ &\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t=\frac{1}{3}+\sqrt{\frac{2}{3}x-\frac{2}{9}}. \\ &\text{代入直线}AB\text{方程中得到} \\ &\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3x^2+9y^2-6x-2\sqrt{3}y-6\sqrt{3}xy+3=0. \\ &\text{这}是\text{一}个\text{非标准的抛物线方程}, \text{将其逆时针旋转}\frac{\pi}{3}\text{即可得到标准的抛物线方程。} \end{align}

\begin{align} \\ &\text{采用如下旋转变换：}\left( \begin{array}{c} x\\ y\\ \end{array} \right) \rightarrow \left( \begin{matrix} \frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& \frac{1}{2}\\ \end{matrix} \right) \left( \begin{array}{c} x\\ y\\ \end{array} \right) =\left( \begin{array}{c} \frac{1}{2}x+\frac{\sqrt{3}}{2}y\\ -\frac{\sqrt{3}}{2}x+\frac{1}{2}y\\ \end{array} \right) , \text{得到} \\ &\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y=\sqrt{3}x^2+\frac{\sqrt{3}}{4}. \\ &\text{这}是\text{标准的抛物线方程。} \\ &\text{根据对称性},\,\text{另外}两个\text{包络线也}是\text{抛物线。}\!\!\text{假设中间的抛物线}为 \\ &\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y=-\sqrt{3}\left( x-\frac{1}{2} \right) ^2+b,\, \\ &\text{代入点}\left( 0,0 \right) ,\,\text{得到}b=\frac{\sqrt{3}}{4}, \text{于}是\text{中间的包络线}为\ \ y=-\sqrt{3}\left( x-\frac{1}{2} \right) ^2+\frac{\sqrt{3}}{4}. \\ &\text{将这}个\text{包络线方程平移变换}为 \\ &\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y=-\sqrt{3}x^2+\frac{\sqrt{3}}{12}. \\ &\text{其极坐标方程}为 r\sin \theta =-\sqrt{3}r^2\cos ^2\theta +\frac{\sqrt{3}}{12}, \text{取}\theta \in \left[ \frac{\pi}{6}, \frac{\pi}{2} \right] , \text{解得} \\ &\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r=\frac{1-\sin \theta}{2\sqrt{3}\cos ^2\theta}. \\ &\text{于}是\text{空白部分面积}为 \\ &\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{d\theta}\int_0^{\frac{1-\sin \theta}{2\sqrt{3}\cos ^2\theta}}{r}dr=3\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\left( \frac{1-\sin \theta}{2\sqrt{3}\cos ^2\theta} \right) ^2d\theta}=\frac{5}{36\sqrt{3}}. \\ &\text{因此}\color{red}{\text{阴影部分面积} S=\frac{\sqrt{3}}{4}-\frac{5}{36\sqrt{3}}=\frac{11}{54}\sqrt{3}}. \\ \end{align}

\begin{align} \\ &\color{orange}{\mathbf{Solution}\,\,\mathbf{3}}\,\,\text{记}F\left( x,y,a \right) =\left( \frac{x}{2-a} \right) ^2+\left( \frac{y}{a} \right) ^2-1, a\in \left( 0,2 \right) . \\ &\text{令}\frac{\partial}{\partial a}F\left( x,y,a \right) =\frac{2x^2}{\left( 2-a \right) ^3}-\frac{2y^2}{a^3}=0, \text{得到}a=\frac{2}{1+\left( \frac{x}{y} \right) ^{\frac{2}{3}}}, \text{代入}F\left( x,y,a \right) =0, \\ &\text{得到包络线}: x^2\frac{\left[ 1+\left( \frac{x}{y} \right) ^{\frac{2}{3}} \right] ^2}{\left( \frac{x}{y} \right) ^{\frac{4}{3}}}+y^2\left[ 1+\left( \frac{x}{y} \right) ^{\frac{2}{3}} \right] ^2=4, \text{化简得到}: \\ &\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{red}{x^{\frac{2}{3}}+y^{\frac{2}{3}}=2^{\frac{2}{3}}}. \\ &\text{这}是\text{星形线。参数方程}为\begin{cases} x=2\cos ^3t\\ y=2\sin ^3t\\ \end{cases}, \text{面积}为 \\ &\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{red}{S=4\int_{\frac{\pi}{2}}^0{2\sin ^3t}\text{d}\left( 2\cos ^3t \right) =48\int_0^{\frac{\pi}{2}}{\cos ^2t\sin ^4t}\text{d}t=\frac{3\pi}{2}}. \\ \end{align}

依旧采用Desmos来验证这题结果的正确性：