Page:Annales de mathématiques pures et appliquées, 1826-1827, Tome 17.djvu/11

$x'\operatorname {d} x'+y'\operatorname {d} y'=0,$

$3r\operatorname {d} x=2(r-2x')\operatorname {d} x',$

$3r\operatorname {d} y=2(r-x')\operatorname {d} y'-2y'\operatorname {d} x'.$

En divisant la troisième par la seconde, il vient

${\frac {\operatorname {d} y}{\operatorname {d} x}}={\frac {(r-x')\operatorname {d} y'-2y'\operatorname {d} x'}{(r-2x')\operatorname {d} x'}}={\frac {(r-x'){\frac {\operatorname {d} y'}{\operatorname {d} x'}}-y'}{r-2x'}}\,;$

ou en mettant pour ${\frac {\operatorname {d} y'}{\operatorname {d} x'}}$ sa valeur tirée de la première

${\frac {\operatorname {d} y}{\operatorname {d} x}}=-{\frac {(r-x')x'+y'^{2}}{(r-2x')y'}}=-{\frac {(r-x')x'+(r^{2}-x'^{2})}{(r-2x')y'}}=-{\frac {(r-x')(r+2x')}{(r-2x')y'}}$

donc

$\left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)^{2}={\frac {(r-x')^{2}(r+2x')^{2}}{(r-2x')^{2}y'^{2}}}={\frac {(r-x')^{2}(r+2x')^{2}}{(r-2x')^{2}\left(r^{2}-x'^{2}\right)}}={\frac {(r-x')(r+2x')^{2}}{(r+x')(r-2x')^{2}}}\,;$

et par suite

$1+\left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)^{2}={\frac {(r+x')(r-2x')^{2}+(r-x')(r+2x')^{2}}{(r+x')(r-2x')^{2}}}={\frac {2r^{3}}{(r+x')(r-2x')^{2}}}.$

D’un autre côté nous venons de trouver

$\operatorname {d} x={\frac {2(r-2x')}{3r}}\operatorname {d} x'\,;$

donc