Page:Annales de mathématiques pures et appliquées, 1825-1826, Tome 16.djvu/16

 Les corrections sont expliquées en page de discussion

$\left.{\begin{aligned}&\left\{\left[(x-a)+p(z-c)\right]+{\frac {n^{2}}{m^{2}}}\left[(a-a')+p(c-c')\right]\right\}{\frac {\operatorname {d} a}{\operatorname {d} a'}}\\\\+&\left\{\left[(y-b)+q(z-c)\right]+{\frac {n^{2}}{m^{2}}}\left[(b-b')+q(c-c')\right]\right\}{\frac {\operatorname {d} b}{\operatorname {d} a'}}\end{aligned}}\right\}$

$={\frac {n^{2}}{m^{2}}}\left\{(a-a')+p'(c-c')\right\},$

$\left.{\begin{aligned}&\left\{\left[(x-a)+p(z-c)\right]+{\frac {n^{2}}{m^{2}}}\left[(a-a')+p(c-c')\right]\right\}{\frac {\operatorname {d} a}{\operatorname {d} b'}}\\\\+&\left\{\left[(y-b)+q(z-c)\right]+{\frac {n^{2}}{m^{2}}}\left[(b-b')+q(c-c')\right]\right\}{\frac {\operatorname {d} b}{\operatorname {d} b'}}\end{aligned}}\right\}$

$={\frac {n^{2}}{m^{2}}}\left\{(b-b')+p'(c-c')\right\}\,;$

ou simplement, en vertu des relations (P’), (Q’)

$\left.{\begin{aligned}&\left\{\left[(x-a)+p(z-c)\right]+{\frac {n^{2}}{m^{2}}}\left[(a-a')+p(c-c')\right]\right\}{\frac {\operatorname {d} a}{\operatorname {d} a'}}\\\\+&\left\{\left[(y-b)+q(z-c)\right]+{\frac {n^{2}}{m^{2}}}\left[(b-b')+q(c-c')\right]\right\}{\frac {\operatorname {d} b}{\operatorname {d} a'}}\end{aligned}}\right\}=0,$

$\left.{\begin{aligned}&\left\{\left[(x-a)+p(z-c)\right]+{\frac {n^{2}}{m^{2}}}\left[(a-a')+p(c-c')\right]\right\}{\frac {\operatorname {d} a}{\operatorname {d} b'}}\\\\+&\left\{\left[(y-b)+q(z-c)\right]+{\frac {n^{2}}{m^{2}}}\left[(b-b')+q(c-c')\right]\right\}{\frac {\operatorname {d} b}{\operatorname {d} b'}}\end{aligned}}\right\}=0\,;$

or, les surfaces (S), (S’) pouvant toutes deux être prises arbitrairement, il s’ensuit que $a',b',c'$ doivent être tout-à-fait indépendans de $a,b,c$ et conséquemment ces deux dernières équations ne doivent établir aucune relation entre les quatre rapports ${\frac {\operatorname {d} a}{\operatorname {d} a'}},\ {\frac {\operatorname {d} b}{\operatorname {d} a'}},\ {\frac {\operatorname {d} a}{\operatorname {d} b'}},\ {\frac {\operatorname {d} b}{\operatorname {d} b'}}\,;$ ce qui exige qu’on ait à la fois

$(x-a)+p(z-c)=-{\frac {n^{2}}{m^{2}}}\left\{(a-a')+p(c-c')\right\}\,;$

$(y-b)+q(z-c)=-{\frac {n^{2}}{m^{2}}}\left\{(b-b')+q(c-c')\right\}\,;$