c’est-à-dire, que les troisièmes sections principales auront aussi la même excentricité, et, par suite, les mêmes foyers. On tirera de là
![{\displaystyle {\frac {1}{A'}}-{\frac {1}{A}}={\frac {1}{B'}}-{\frac {1}{B}}={\frac {1}{C'}}-{\frac {1}{C}},\quad {\frac {1}{A''}}-{\frac {1}{A}}={\frac {1}{B''}}-{\frac {1}{B}}={\frac {1}{C''}}-{\frac {1}{C}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e9ddd86a2a9aca1ce20f7f5c03b8289c3fabb4c)
(12)
au moyen de quoi on aura
![{\displaystyle {\frac {a'^{2}}{A'}}+{\frac {b'^{2}}{B'}}+{\frac {c'^{2}}{C'}}={\frac {a'^{2}}{A}}+{\frac {b'^{2}}{B}}+{\frac {c'^{2}}{C}}+\left({\frac {1}{A'}}-{\frac {1}{A}}\right)a'^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/883bef234efbb21c832a7c8d86c3c328b0ad52b6)
![{\displaystyle +\left({\frac {1}{B'}}-{\frac {1}{B}}\right)b'^{2}+\left({\frac {1}{C'}}-{\frac {1}{C}}\right)c'^{2},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13cea23ea2440424a2c19a23a0b2f259fbf82e82)
![{\displaystyle {\frac {a''^{2}}{A''}}+{\frac {b''^{2}}{B''}}+{\frac {c''^{2}}{C''}}={\frac {a''^{2}}{A}}+{\frac {b''^{2}}{B}}+{\frac {c''^{2}}{C}}+\left({\frac {1}{A''}}-{\frac {1}{A}}\right)a''^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ff9ae927c37663a5d429d6d7aa544593d4f8875)
![{\displaystyle +\left({\frac {1}{B''}}-{\frac {1}{B}}\right)b''^{2}+\left({\frac {1}{C''}}-{\frac {1}{C}}\right)c''^{2},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/562f8b1360ac1aba28c3e36ee205994a708d6b1d)
c’est-à-dire (5), (6), (12),
![{\displaystyle {\begin{aligned}{\frac {a'^{2}}{A}}+\,{\frac {b'^{2}}{B}}+\,{\frac {c'^{2}}{C}}\,&=1+\left({\frac {1}{B}}-{\frac {1}{B'}}\right)r'^{2},\\\\{\frac {a''^{2}}{A}}+{\frac {b''^{2}}{B}}+{\frac {c''^{2}}{C}}&=1+\left({\frac {1}{C}}-{\frac {1}{C''}}\right)r''^{2}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/382ee1d5634654e38756ca07bc9e7969142c660f)
ou bien encore
![{\displaystyle {\begin{aligned}&{\frac {1}{A}}\left({\frac {a'}{r'}}\right)^{2}\,+{\frac {1}{B}}\left({\frac {b'}{r'}}\right)^{2}\,+{\frac {1}{C}}\left({\frac {c'}{r'}}\right)^{2}\,={\frac {1}{r'^{2}}}+\left({\frac {1}{B}}-{\frac {1}{B'}}\right),\\\\&{\frac {1}{A}}\left({\frac {a''}{r''}}\right)^{2}+{\frac {1}{B}}\left({\frac {b''}{r''}}\right)^{2}+{\frac {1}{C}}\left({\frac {c''}{r''}}\right)^{2}={\frac {1}{r''^{2}}}+\left({\frac {1}{C}}-{\frac {1}{C''}}\right)\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5538dbcae85a0cca136c057c4b115a65d6747611)
mais on a aussi (6)
![{\displaystyle {\frac {1}{A}}\left({\frac {a}{r}}\right)^{2}+{\frac {1}{B}}\left({\frac {b}{r}}\right)^{2}+{\frac {1}{C}}\left({\frac {c}{r}}\right)^{2}={\frac {1}{r^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/439eec3d3fe98609faeeabfdbeda25905a3ce947)
ajoutant ces trois équations membre à membre, en ayant égard aux relations (10), il viendra