Wyznaczamy współrzędne kilku punktów leżących na hiperboli  $y=\frac{4,5}{x}.$  

$x$	$\frac{1}{4}$	$\frac{1}{2}$	$1$
$y$	$18$	$9$	$\frac{9}{2}$

Przyjmijmy, że:

$A=\left(\frac{1}{4},18\right)$  

$B=\left(\frac{1}{2},9\right)$  

$C=\left(1,\frac{9}{2}\right)$  

Sprawdzamy, czy spełnione są warunki danej definicji.

$\left|\left|A{F}_1\right|-\right|A{F}_2\text{||}=\left|\sqrt{{\left(3-\frac{1}{4}\right)}^2+{\left(3-18\right)}^2}-\sqrt{{\left(-3-\frac{1}{4}\right)}^2+{\left(-3-18\right)}^2}\right|=\left|\sqrt{{\left(2\frac{3}{4}\right)}^2+{\left(-15\right)}^2}-\sqrt{{\left(-3\frac{1}{4}\right)}^2+{\left(-21\right)}^2}\right|=\left|\sqrt{{\left(\frac{11}{4}\right)}^2+225}-\sqrt{{\left(-\frac{13}{4}\right)}^2+441}\right|=\left|\sqrt{\frac{121}{16}+\frac{3600}{16}}-\sqrt{\frac{169}{16}+\frac{7056}{16}}\right|=\left|\sqrt{\frac{3721}{16}}-\sqrt{\frac{7225}{16}}\right|=\left|\frac{61}{4}-\frac{85}{4}\right|=\left|-\frac{24}{4}\right|=\left|-6\right|=6$  

$\left|\left|B{F}_1\right|-\right|B{F}_2\text{||}=\left|\sqrt{{\left(3-\frac{1}{2}\right)}^2+{\left(3-9\right)}^2}-\sqrt{{\left(-3-\frac{1}{2}\right)}^2+{\left(-3-9\right)}^2}\right|=\left|\sqrt{{\left(2\frac{1}{2}\right)}^2+{\left(-6\right)}^2}-\sqrt{{\left(-3\frac{1}{2}\right)}^2+{\left(-12\right)}^2}\right|=\left|\sqrt{{\left(\frac{5}{2}\right)}^2+36}-\sqrt{{\left(-\frac{7}{2}\right)}^2+144}\right|=\left|\sqrt{\frac{25}{4}+\frac{144}{4}}-\sqrt{\frac{49}{4}+\frac{576}{4}}\right|=\left|\sqrt{\frac{169}{4}}-\sqrt{\frac{625}{4}}\right|=\left|\frac{13}{2}-\frac{25}{2}\right|=\left|-\frac{12}{2}\right|=\left|-6\right|=6$  

$\left|\left|C{F}_1\right|-\right|C{F}_2\text{||}=\left|\sqrt{{\left(3-1\right)}^2+{\left(3-\frac{9}{2}\right)}^2}-\sqrt{{\left(-3-1\right)}^2+{\left(-3-\frac{9}{2}\right)}^2}\right|=\left|\sqrt{2^2+{\left(-\frac{3}{2}\right)}^2}-\sqrt{{\left(-4\right)}^2+{\left(-\frac{15}{2}\right)}^2}\right|=\left|\sqrt{4+\frac{9}{4}}-\sqrt{16+\frac{225}{4}}\right|=\left|\sqrt{\frac{16}{4}+\frac{9}{4}}-\sqrt{\frac{64}{4}+\frac{225}{4}}\right|=\left|\sqrt{\frac{25}{4}}-\sqrt{\frac{289}{4}}\right|=\left|\frac{5}{2}-\frac{17}{2}\right|=\left|-\frac{12}{2}\right|=\left|-6\right|=6$  

Możemy zauważyć, że:

$\left|\left|A{F}_1\right|-\right|A{F}_2\left|\left|=\right|\right|B{F}_1\left|-\right|B{F}_2\left|\left|=\right|\right|C{F}_1\left|-\right|C{F}_2\mid \mid =6$