2次曲線：有心の場合の標準化の例

2次曲線の標準化に従って，与式

の係数より

であり，これらを成分とする2つの対称行列

${\displaystyle A=\left(\begin{array}{cc}5& -3\\ -3& 5\end{array}\right)}$     ······ 

${\displaystyle \tilde{A}=\left(\begin{array}{ccc}5& -3& 1\\ -3& 5& -7\\ 1& -7& -3\end{array}\right)}$     ······ 

を考える．判別式

${\displaystyle \Delta =\left|A\right|=5\cdot 5-{(-3)}^2}$ ${\displaystyle =16\ne 0}$     ······ 

および，

より，この2次曲線は有心で退化せず， ${\displaystyle \frac{\tilde{\Delta }}{\Delta }=\frac{-256}{16}}$ ${\displaystyle =-16}$ である．また，行列 ${\displaystyle A}$ の固有方程式

${\displaystyle |A-\lambda E|=\left|\begin{array}{cc}5-\lambda & -3\\ -3& 5-\lambda \end{array}\right|}$ ${\displaystyle =(5-\lambda )(5-\lambda )-{(-3)}^2}$ ${\displaystyle ={\lambda }^2-10\lambda +16}$ ${\displaystyle =(\lambda -2)(\lambda -8)}$ ${\displaystyle =0}$

より， ${\displaystyle A}$ の固有値は ${\displaystyle {\lambda }_1=2}$ , ${\displaystyle {\lambda }_2=8}$ である．したがって，2次曲線の標準化の【定理 1】より，与式の2次曲線は適当な座標変換 ${\displaystyle (x,y)\mapsto (X,Y)}$ で

${\displaystyle 2X^2+8Y^2-16=0}$    ${\displaystyle \iff }$    ${\displaystyle \frac{X^2}{8}+\frac{Y^2}{2}=1}$     ······ 

と変換できる．これは，軌道長半径 ${\displaystyle 2\sqrt{2}}$ ， 軌道短半径 ${\displaystyle \sqrt{2}}$ の楕円を表す．

与式の中心の座標 ${\displaystyle (x_0,y_0)}$ は逆行列 ${\displaystyle A^{-1}=\frac{1}{16}\left(\begin{array}{cc}5& 3\\ 3& 5\end{array}\right)}$ を用いて

${\displaystyle \left(\begin{array}{c}{x}_0\\ y_0\end{array}\right)=-A^{-1}\left(\begin{array}{c}1\\ -7\end{array}\right)}$ ${\displaystyle =-\frac{1}{16}\left(\begin{array}{cc}5& 3\\ 3& 5\end{array}\right)\left(\begin{array}{c}1\\ -7\end{array}\right)}$ ${\displaystyle =\left(\begin{array}{c}1\\ 2\end{array}\right)}$     ······ 

と求まり，行列 ${\displaystyle A}$ の固有値 ${\displaystyle {\lambda }_1=2}$ , ${\displaystyle {\lambda }_2=8}$ に対応する（大きさ ${\displaystyle 1}$ の）固有ベクトル ${\displaystyle {\mathit{p}}_1}$ , ${\displaystyle {\mathit{p}}_2}$ は

${\displaystyle \left(\begin{array}{cc}5-2& -3\\ -3& 5-2\end{array}\right){\mathit{p}}_1}$ ${\displaystyle =\left(\begin{array}{cc}3& -3\\ -3& 3\end{array}\right){\mathit{p}}_1}$ ${\displaystyle =0}$    ${\displaystyle \Rightarrow }$    ${\displaystyle {\mathit{p}}_1=\pm \frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)}$     ······ 

${\displaystyle \left(\begin{array}{cc}5-8& -3\\ -3& 5-8\end{array}\right){\mathit{p}}_2}$ ${\displaystyle =\left(\begin{array}{cc}-3& -3\\ -3& -3\end{array}\right){\mathit{p}}_2}$ ${\displaystyle =0}$    ${\displaystyle \Rightarrow }$    ${\displaystyle {\mathit{p}}_2=\pm \frac{1}{\sqrt{2}}\left(\begin{array}{c}-1\\ 1\end{array}\right)}$     ······ 

と求まる．よって， ${\displaystyle A}$ の対角化を与える（回転変換に対応させた）直交行列を

${\displaystyle P=\left(\begin{array}{cc}{\mathit{p}}_1& {\mathit{p}}_2\end{array}\right)}$ ${\displaystyle =\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)}$ ${\displaystyle =\left(\begin{array}{cc}\cos 45°& -\sin 45°\\ \sin 45°& \cos 45°\end{array}\right)}$     ······ 

と書けるので， ${\displaystyle (X,Y)\mapsto (x,y)}$ の変換は

${\displaystyle \left(\begin{array}{c}x\\ y\end{array}\right)=P\left(\begin{array}{c}X\\ Y\end{array}\right)+\left(\begin{array}{c}{x}_0\\ y_0\end{array}\right)}$ ${\displaystyle =\left(\begin{array}{cc}\cos 45°& -\sin 45°\\ \sin 45°& \cos 45°\end{array}\right)\left(\begin{array}{c}X\\ Y\end{array}\right)+\left(\begin{array}{c}1\\ 2\end{array}\right)}$     ······ 

と表せる．以上のことから，与式の2次曲線は，式の楕円を，原点を中心に ${\displaystyle 45°}$ 回転させた後，楕円の中心を ${\displaystyle (1,2)}$ まで平行移動させたものである．

楕円の回転と平行移動 JSXGraph Copyright (C) see http://jsxgraph.org