Page:Annales de mathématiques pures et appliquées, 1829-1830, Tome 20.djvu/173

$P={\frac {\operatorname {d} S}{\operatorname {d} x}}=2x,\quad Q={\frac {\operatorname {d} S}{\operatorname {d} y}}=2y,\quad P^{2}+Q^{2}=4\left(x^{2}+y^{2}\right)=4r^{2}\,;$

ce qui donnera, en substituant dans (23), et observant qu’ici $A=a$

$rN=(z+a)\left(x{\frac {\operatorname {d} ^{2}x}{\operatorname {d} s^{2}}}+y{\frac {\operatorname {d} ^{2}y}{\operatorname {d} s^{2}}}\right)\,;\qquad$(27)

mais, par deux différentiations successives, l’équation (26) donne

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

$x{\frac {\operatorname {d} ^{2}x}{\operatorname {d} s^{2}}}+y{\frac {\operatorname {d} ^{2}y}{\operatorname {d} s^{2}}}=-\left({\frac {\operatorname {d} x}{\operatorname {d} s}}\right)^{2}-\left({\frac {\operatorname {d} y}{\operatorname {d} s}}\right)^{2}=\left({\frac {\operatorname {d} z}{\operatorname {d} s}}\right)^{2}-1\,;$

on aura donc, en substituant dans (27),

$rN=(z+a)\left\{\left({\frac {\operatorname {d} z}{\operatorname {d} s}}\right)^{2}-1\right\}\,;$

d’un autre côté, l’équation (25) donne

${\frac {\operatorname {d} z}{\operatorname {d} s}}={\frac {1}{2}}\left(e^{\frac {s}{a}}-e^{-{\frac {s}{a}}}\right),$

d’où

$\left({\frac {\operatorname {d} z}{\operatorname {d} s}}\right)^{2}={\frac {1}{4}}\left(e^{\frac {2s}{a}}-e^{-{\frac {2s}{a}}}-2\right),$

et

$\left({\frac {\operatorname {d} z}{\operatorname {d} s}}\right)^{2}-1={\frac {1}{4}}\left(e^{\frac {2s}{a}}+e^{-{\frac {2s}{a}}}+2\right)=\left({\frac {e^{\frac {s}{a}}+e^{-{\frac {s}{a}}}}{2}}\right)^{2}=\left({\frac {z+a}{a}}\right)^{2}\,;$

il viendra donc, en substituant,