Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/40

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INTÉGRALES

${\frac {\left(1+y'^{2}\right)x''-x'y'y''}{(1+x'^{2}+y'^{2})^{\frac {3}{2}}}}=0,\qquad {\frac {\left(1+x'^{2}\right)y''-x'y'x''}{(1+x'^{2}+y'^{2})^{\frac {3}{2}}}}=0,$

qu’on pourra mettre ensuite sous cette forme

${\frac {\left(1+x'^{2}+y'^{2}\right)x''-(x'x''+y'y'')x'}{(1+x'^{2}+y'^{2})^{\frac {3}{2}}}}=0,\ {\frac {\left(1+x'^{2}+y'^{2}\right)y''-(x'x''+y'y'')y'}{(1+x'^{2}+y'^{2})^{\frac {3}{2}}}}=0,$

ou, en continuant d’employer les notations de Lagrange,

${\frac {\left(1+x'^{2}+y'^{2}\right)x''-\left(1+x'^{2}+y'^{2}\right)'x'}{(1+x'^{2}+y'^{2})^{\frac {3}{2}}}}=0,\ {\frac {\left(1+x'^{2}+y'^{2}\right)y''-\left(1+x'^{2}+y'^{2}\right)'y'}{(1+x'^{2}+y'^{2})^{\frac {3}{2}}}}=0,$

ou encore

${\frac {x''{\sqrt {1+x'^{2}+y'^{2}}}-x'\left({\sqrt {1+x'^{2}+y'^{2}}}\right)'}{1+x'^{2}+y'^{2}}}=0,\ {\frac {y''{\sqrt {1+x'^{2}+y'^{2}}}-y'\left({\sqrt {1+x'^{2}+y'^{2}}}\right)'}{1+x'^{2}+y'^{2}}}=0,$

ou enfin

$\left({\frac {x'}{\sqrt {1+x'^{2}+y'^{2}}}}\right)'=0,\qquad \left({\frac {y'}{\sqrt {1+x'^{2}+y'^{2}}}}\right)'=0,$

ce qui donne

${\frac {x'}{\sqrt {1+x'^{2}+y'^{2}}}}=A,\qquad {\frac {y'}{\sqrt {1+x'^{2}+y'^{2}}}}=B.$

En considérant, dans ces équations, $x'$ et $y'$ comme deux inconnues, on en tire, en transformant les constantes,

$x'={\frac {A}{\sqrt {1-A^{2}-B^{2}}}}=M,\qquad y'={\frac {B}{\sqrt {1-A^{2}-B^{2}}}}=N,$

d’où enfin

$x=Mz+G,\qquad y=Nz+H,$