${\displaystyle \frac{1}{x}}$のグラフ

${\displaystyle \left|{\text{F}}_1\text{P}-{\text{F}}_2\text{P}\right|}$${\displaystyle =\left|\sqrt{{\left(r-\sqrt{2}\right)}^2+{\left(\frac{1}{r}-\sqrt{2}\right)}^2}-\sqrt{{\left(r+\sqrt{2}\right)}^2+{\left(\frac{1}{r}+\sqrt{2}\right)}^2}\right|}$

${\displaystyle =\left|\sqrt{r^2-2\sqrt{2}r+2+\frac{1}{r^2}-\frac{2\sqrt{2}}{r}+2}-\sqrt{r^2+2\sqrt{2}r+2+\frac{1}{r^2}+\frac{2\sqrt{2}}{r}+2}\right|}$

${\displaystyle =\left|\sqrt{r^2+2+\frac{1}{r^2}-2\sqrt{2}\left(r+\frac{1}{r}\right)+2}-\sqrt{r^2+2+\frac{1}{r^2}+2\sqrt{2}\left(r+\frac{1}{r}\right)+2}\right|}$

${\displaystyle =\left|\sqrt{{\left(r+\frac{1}{r}\right)}^2-2\sqrt{2}\left(r+\frac{1}{r}\right)+2}-\sqrt{{\left(r+\frac{1}{r}\right)}^2+2\sqrt{2}\left(r+\frac{1}{r}\right)+2}\right|}$

${\displaystyle ==\left|\sqrt{{\left\{\left(r+\frac{1}{r}\right)-\sqrt{2}\right\}}^2}-\sqrt{{\left\{\left(r+\frac{1}{r}\right)+\sqrt{2}\right\}}^2}\right|}$

${\displaystyle =\left|\left\{\left(r+\frac{1}{r}\right)-\sqrt{2}\right\}-\left\{\left(r+\frac{1}{r}\right)+\sqrt{2}\right\}\right|}$

${\displaystyle =\left|-2\sqrt{2}\right|}$

${\displaystyle =2\sqrt{2}}$　･･････(2)

${\displaystyle \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}\cos \left(-45°\right)& -\sin \left(-45°\right)\\ \sin \left(-45°\right)& \cos \left(-45°\right)\end{array}\right)\left(\begin{array}{c}r\\ s\end{array}\right)}$

${\displaystyle \left(\begin{array}{c}r\\ s\end{array}\right)=\left(\begin{array}{cc}\cos 45°& -\sin 45°\\ \sin 45°& \cos 45°\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)}$

${\displaystyle \frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y=\frac{1}{\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}y}}$

${\displaystyle \left(\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y\right)\left(\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}y\right)=1}$