Page:Annales de mathématiques pures et appliquées, 1818-1819, Tome 9.djvu/145

139

ET DES SURFACES COURBES.

${\frac {(x'+x)^{2}}{a^{2}}}+{\frac {(y'+y)^{2}}{b^{2}}}+{\frac {(z'+z)^{2}}{c^{2}}}=1,$

${\frac {(x'+x)^{2}}{a'^{2}}}+{\frac {(y'+y)^{2}}{b'^{2}}}+{\frac {(z'+z)^{2}}{c'^{2}}}=1\,;$

puis, en développant ;

$0=\left({\frac {x'^{2}}{a^{2}}}+{\frac {y'^{2}}{b^{2}}}+{\frac {z'^{2}}{c^{2}}}-1\right)+{\frac {2x'x}{a^{2}}}+{\frac {2y'y}{b^{2}}}+{\frac {z'z}{c^{2}}}+\ldots$

$0=\left({\frac {x'^{2}}{a'^{2}}}+{\frac {y'^{2}}{b'^{2}}}+{\frac {z'^{2}}{c'^{2}}}-1\right)+{\frac {2x'x}{a'^{2}}}+{\frac {2y'y}{b'^{2}}}+{\frac {z'z}{c'^{2}}}+\ldots$

Parce que le point $(x',y',z')$ est sur la courbe, nous aurons d’abord les deux équations de condition.

${\frac {x'^{2}}{a^{2}}}+{\frac {y'^{2}}{b^{2}}}+{\frac {z'^{2}}{c^{2}}}=1,\qquad$(5)

${\frac {x'^{2}}{a'^{2}}}+{\frac {y'^{2}}{b'^{2}}}+{\frac {z'^{2}}{c'^{2}}}=1\,;\qquad$(5′)

et les équations de la tangente, rapportée aux axes primitifs seront

${\frac {x'(x-x')}{a^{2}}}+{\frac {y'(y-y')}{b^{2}}}+{\frac {z'(z-z')}{c^{2}}}=0,$

${\frac {x'(x-x')}{a'^{2}}}+{\frac {y'(y-y')}{b'^{2}}}+{\frac {z'(z-z')}{c'^{2}}}=0\,;$

ou simplement, en vertu des conditions (5, 5′)

${\frac {x'x}{a^{2}}}+{\frac {y'y}{b^{2}}}+{\frac {z'z}{c^{2}}}=1.\qquad$(6)

${\frac {x'x}{a'^{2}}}+{\frac {y'y}{b'^{2}}}+{\frac {z'z}{c'^{2}}}=1.\qquad$(6′)