C..1.1.1 証明

$\displaystyle \sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a.$

$\displaystyle \sqrt{(x-c)^2+y^2}=2a-\sqrt{(x+c)^2+y^2}$

$\displaystyle (x-c)^2+y^2=4a^2-4a\sqrt{(x+c)^2+y^2}+{(x+c)^2+y^2}$

$\displaystyle cx+a^2=a\sqrt{(x+c)^2+y^2}$

$\displaystyle c^2x^2+2a^2cx+a^4=a^2\left[(x+c)^2+y^2\right]$

$\displaystyle (a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2).$

$b:=\sqrt{a^2-c^2}$ とおくと、

$\displaystyle b^2x^2+a^2y^2=a^2b^2.$

$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. \qed$

桂田祐史