46303 再談等角差線－兼談108數學課綱之圓錐曲線教學

1. 有趣的問題，可賀的恆心

\begin{align} x \tan(\theta +\alpha )=y=(\ell-x) \tan\theta,\label{1} \end{align}

\begin{align} \left\{ \begin{array}{l} x=\dfrac{\ell \tan\theta}{\tan(\theta +\alpha)+\tan\theta},\\[12pt] y=\dfrac{\ell \tan\theta\tan( \theta +\alpha)}{\tan( \theta +\alpha)+\tan\theta}.\end{array}\right.\label{2} \end{align}

2. 檢視圖 2 與原問題的關係

\begin{align*} \lim\limits_{\theta\rightarrow \frac\pi{3}}\frac{\tan\theta}{\tan(\theta+\frac{\pi}{6})+\tan\theta}=\,&0,\\ \lim\limits_{\theta\rightarrow \frac\pi{3}}\frac{\tan\theta\tan(\theta+\frac{\pi}{6})}{\tan(\theta+\frac{\pi}{6})+\tan\theta}=\,&\tan \frac{\pi}3=\sqrt{3}. \end{align*}

3. 後見之明 --- 看起來像是雙曲線

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$

$$x=a/\cos\theta\quad \hbox{、}\quad y=b\tan\theta.$$

$$\frac yx=\tan(\theta+\alpha)\quad \hbox{、}\quad \frac y{\ell-x}=\tan\theta,$$

$$\frac yx=\tan(\theta+\alpha)=\frac{\tan\theta+\tan\alpha}{1-\tan\theta\tan\alpha}=\dfrac{\dfrac{y}{\ell-x}+\tan\alpha}{1-\dfrac{y}{\ell-x}\tan\alpha} =\frac{y+(\ell-x)\tan\alpha}{(\ell-x)-y\tan\alpha},$$

\begin{align} x^2\tan\alpha-2xy-y^2\tan\alpha-x\ell\tan\alpha+y\ell=0.\label{3} \end{align}

$$(x+\frac{\ell}2)^2\tan\alpha-2(x+\frac{\ell}2)y-y^2\tan\alpha-(x+\frac{\ell}2)\ell\tan\alpha+y\ell=0,$$

\begin{align} x^2\tan\alpha-2xy-y^2\tan\alpha=\frac{\ell^2\tan\alpha}{4}.\label{4} \end{align}

$$\left[\begin{array}{c} x'\\ y'\end{array}\right]= \left[\begin{array}{ccc} \cos\beta&~\quad&-\sin\beta\\ \sin\beta&&\cos\beta\end{array}\right] \left[\begin{array}{c} x\\ y\end{array}\right],$$

$$\left[\begin{array}{c} x\\ y\end{array}\right]= \left[\begin{array}{ccc} \cos\beta&~\quad&\sin\beta\\ -\sin\beta&&\cos\beta\end{array}\right]\left[\begin{array}{c} x'\\ y'\end{array}\right]=\left[\begin{array}{c} x'\cos\beta+y'\sin\beta\\ -x'\sin\beta+y'\cos\beta\end{array}\right].$$

$$(x^2-y^2)(\cos2\beta\tan\alpha+\sin2\beta)+2xy(\sin2\beta\tan\alpha-\cos2\beta)=\frac{\ell^2\tan\alpha}{4},$$

$$\cos2\beta\tan\alpha+\sin2\beta=\sin\alpha\tan\alpha+\cos\alpha=\frac 1{\cos\alpha},$$

\begin{align} x^2-y^2=\frac{\ell^2\sin\alpha}{4}.\label{5} \end{align}

$$\sin2\beta\tan\alpha-\cos2\beta=\sin\alpha\tan\alpha+\cos\alpha=\frac 1{\cos\alpha},$$

\begin{align} xy=\frac{\ell^2\sin\alpha}{4}.\label{6} \end{align}

$$x^2\tan\frac \alpha 2+xy(\tan^2\frac \alpha 2-1)-y^2\tan\frac \alpha 2=0,$$

4. 早期高中的圓錐曲線教學

4.1. 用焦點與準線定義圓錐曲線

$$\sqrt{(x-h)^2+(y-k)^2}=\frac{e|ax+by+c|}{\sqrt{a^2+b^2}},$$

$$(1-e^2)x^2+y^2-(2h+2e^2c)x+(h^2-e^2c^2)=0.$$

$$y^2=4cx,$$

$$(1-e^2)x^2+y^2+e^2(e^2-1)c^2=0;$$

$$\dfrac{x^2}{{e^2c^2}}+\frac{y^2}{e^2(1-e^2)c^2}=1;$$

$$\dfrac{x^2}{{e^2c^2}}-\frac{y^2}{e^2(e^2-1)c^2}=1.$$

4.2. 二元二次方程式的平移

\begin{align} ax^2+bxy+cy^2+dx+ey+f=0.\label{7} \end{align}

$$a(x-h)^2+b(x-h)(y-k)+c(y-k)^2+d(x-h)+e(y-k)+f=0.$$

\begin{align} \left\{\begin{array}{l} 2ah+bk=d,\\[5pt] bh+2ck=e. \end{array}\right.\label{8} \end{align}

\begin{align} ax^2+bxy+cy^2+f'=0.\label{9} \end{align}

4.3. 二元二次方程式的旋轉

\begin{align*} a(x\cos\beta+y\sin\beta)^2+b(x\cos\beta+y\sin\beta)(-x\sin\beta+y\cos\beta)+c(-x\sin\beta+y\cos\beta)^2\\ +d(x\cos\beta+y\sin\beta)+e(-x\sin\beta+y\cos\beta)+f=0, \end{align*}

\begin{align*} A=\,&a\cos^2\beta-b\sin\beta\cos\beta+c\sin^2\beta,\\ B=\,&2a\sin\beta\cos\beta+b(\cos^2\beta-\sin^2\beta)-2c\sin\beta\cos\beta\\ =\,&(a-c)\sin2\beta+b\cos2\beta,\\ C=\,&a\sin^2\beta+b\sin\beta\cos\beta+c\cos^2\beta, \end{align*}

$$\tan2\beta=\frac b{c-a},$$

\begin{align} A+C=\,&a(\cos^2\beta+\sin^2\beta)+c(\cos^2\beta+\sin^2\beta)=a+c,\label{10}\\ A-C=\,&(a-c)(\cos^2\beta-\sin^2\beta)-2b\sin\beta\cos\beta\nonumber\\ =\,&(a-c)\cos2\beta-b\sin2\beta,\nonumber\\ (A-C)^2+&B^2=(a-c)^2(\cos^22\beta+\sin^22\beta)+b^2(\sin^2\beta+\cos^22\beta)\nonumber\\ &\quad\, =(a-c)^2+b^2.\label{11} \end{align}

\begin{align} B^2-4AC=b^2-4ac.\label{12} \end{align}