2次曲線：無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

${\displaystyle A=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)}$     ······ 

${\displaystyle \tilde{A}=\left(\begin{array}{ccc}1& -1& 2\\ -1& 1& 2\\ 2& 2& -8\end{array}\right)}$     ······ 

を考える．判別式

${\displaystyle \Delta =\left|A\right|=1\cdot 1-{(-1)}^2}$ ${\displaystyle =0}$     ······ 

および，

より，この2次曲線は無心で退化しない．また，行列 ${\displaystyle A}$ の固有値は ${\displaystyle {\lambda }_1=0}$ , ${\displaystyle {\lambda }_2=a+c}$ ${\displaystyle =2}$ であり，

${\displaystyle \left(\begin{array}{c}{x}_0\\ y_0\end{array}\right)=-\frac{1}{{\lambda }_2}\left(\begin{array}{c}d\\ e\end{array}\right)}$ ${\displaystyle =-\frac{1}{2}\left(\begin{array}{c}2\\ 2\end{array}\right)}$ ${\displaystyle =\left(\begin{array}{c}-1\\ -1\end{array}\right)}$     ······ 

より，

である．さらに，

${\displaystyle g=\frac{ae-bd}{\sqrt{a^2+b^2}}}$ ${\displaystyle =\frac{1\cdot 2-(-1)\cdot 2}{\sqrt{1^2+1^2}}}$ ${\displaystyle =\frac{4}{\sqrt{2}}}$ ${\displaystyle =2\sqrt{2}}$     ······ 

であり，2次曲線の標準化の【定理 2】より，与式の2次曲線は適当な座標変換 ${\displaystyle (x,y)\mapsto (X,Y)}$ で

${\displaystyle 2Y^2-4\sqrt{2}X-16=0}$    ${\displaystyle \iff }$    ${\displaystyle Y^2=2\sqrt{2}(X+2\sqrt{2})}$     ······ 

と変換できる．これは，頂点 ${\displaystyle (-2\sqrt{2},0)}$ の放物線を表す．

行列 ${\displaystyle A}$ の固有値 ${\displaystyle {\lambda }_1=0}$ , ${\displaystyle {\lambda }_2=2}$ 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

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

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

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

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

と表せる．以上のことから，与式の2次曲線は，式の放物線を，原点を中心に ${\displaystyle 45°}$ 回転させた後，その原点を ${\displaystyle (-1,-1)}$ まで平行移動させたものである．