Page:Annales de mathématiques pures et appliquées, 1811-1812, Tome 2.djvu/80

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CENTRE DES MOYENNES DISTANCES

Première supposition. Soit, $c=a,$ on aura

${\frac {\operatorname {d} z}{\operatorname {d} y}}={\frac {2}{(a+c)+2by}}={\frac {1}{b}}\cdot {\frac {2b}{(a+c)+2by}}={\frac {1}{b}}\cdot {\frac {2b}{({\frac {a+c}{2b}})+y}}$

d’où $z=\mathrm {C} +{\frac {1}{b}}Log.\left\{{\frac {a+c}{2b}}+\operatorname {Tang} .{\tfrac {1}{2}}x\right\}.$

Si, en particulier, on suppose que $z$ et $x$ doivent être nuls en même temps, on aura

$z={\frac {1}{b}}Log.\left\{1+{\frac {2b}{a+c}}\operatorname {Tang} .{\tfrac {1}{2}}x\right\}.$

Seconde supposition. Soit $c>a$, on aura

${\begin{aligned}{\frac {\operatorname {d} z}{\operatorname {d} x}}&=2\cdot {\frac {1}{(c+a)+2by-(c-a)y^{2}}}\\&={\frac {2}{(c-a)}}\cdot {\frac {1}{{\frac {(c+a)}{(c-a)}}+{\frac {2b}{(c-a)}}-y^{2}}}\\&={\frac {2}{(c-a)}}\cdot {\frac {1}{\left\{{\frac {(c+a)}{(c-a)}}+{\frac {b^{2}}{(c-a)^{2}}}\right\}-\left\{{\frac {b}{c-a}}-y\right\}^{2}}}\\&={\frac {2}{(c-a)}}\cdot {\frac {1}{{\frac {c^{2}-a^{2}+b^{2}}{(c-a)^{2}}}-\left\{{\frac {b}{c-a}}-y\right\}^{2}}}\\&={\frac {2}{(c-a)}}\cdot {\frac {1}{\left\{{\frac {{\sqrt {(c^{2}-a^{2}+b^{2})}}+b}{(c-a)}}-y\right\}\left\{{\frac {{\sqrt {(c^{2}-a^{2}+b^{2})}}-b}{(c-a)}}+y\right\}}}\\&={\frac {1}{\sqrt {c^{2}-a^{2}+b^{2}}}}\left\{{\frac {1}{{\frac {\sqrt {c^{2}-a^{2}+b^{2}}}{c-a}}-y}}+{\frac {1}{{\frac {\sqrt {c^{2}-a^{2}+b^{2}}}{c-a}}+y}}\right\}\\\end{aligned}}$

d’où on conclura

$z=\mathrm {C} +{\tfrac {1}{\sqrt {c^{2}-a^{2}+b^{2}}}}\left\{Log.\left[{\tfrac {\sqrt {c^{2}-a^{2}+b^{2}}}{c-a}}+y\right]-Log.\left[{\tfrac {\sqrt {c^{2}-a^{2}+b^{2}}}{c-a}}-y\right]\right\}$