Page:Annales de mathématiques pures et appliquées, 1825-1826, Tome 16.djvu/94

${\begin{array}{lr}{\frac {1}{2}}(\alpha +t'),&{\frac {1}{2}}(\alpha '+t),\\\\{\frac {1}{2}}(\beta +u'),&{\frac {1}{2}}(\beta '+u).\end{array}}$

En conséquence, la droite $\mathrm {Q,Q'}$ fera avec l’axe des $x$ un angle $n$ dont la tangente tabulaire sera

$\operatorname {Tang} .n={\frac {(\beta '+u)-(\beta +u')}{(\alpha '+t)-(\alpha +t')}}={\frac {(u-\beta )-(u'-\beta ')}{(t-\alpha )-(t'-\alpha ')}}.$

Or on a

$t=\alpha -{\frac {1+\left({\frac {\operatorname {d} \beta }{\operatorname {d} \alpha }}\right)^{2}}{\frac {\operatorname {d} ^{2}\beta }{\operatorname {d} \alpha ^{2}}}}{\frac {\operatorname {d} \beta }{\operatorname {d} \alpha }},\qquad u=\beta +{\frac {1+\left({\frac {\operatorname {d} \beta }{\operatorname {d} \alpha }}\right)^{2}}{\frac {\operatorname {d} ^{2}\beta }{\operatorname {d} \alpha ^{2}}}},$

$t'=\alpha '-{\frac {1+\left({\frac {\operatorname {d} \beta '}{\operatorname {d} \alpha '}}\right)^{2}}{\frac {\operatorname {d} ^{2}\beta '}{\operatorname {d} \alpha '^{2}}}}{\frac {\operatorname {d} \beta '}{\operatorname {d} \alpha '}},\qquad u'=\beta '+{\frac {1+\left({\frac {\operatorname {d} \beta '}{\operatorname {d} \alpha '}}\right)^{2}}{\frac {\operatorname {d} ^{2}\beta '}{\operatorname {d} \alpha '^{2}}}}\,;$

valeurs qui, substituées dans celle de $\operatorname {Tang} .n,$ donnent

$\operatorname {Tang} .n=-{\frac {{\frac {1+\left({\frac {\operatorname {d} \beta }{\operatorname {d} \alpha }}\right)^{2}}{\frac {\operatorname {d} ^{2}\beta }{\operatorname {d} \alpha ^{2}}}}-{\frac {1+\left({\frac {\operatorname {d} \beta '}{\operatorname {d} \alpha '}}\right)^{2}}{\frac {\operatorname {d} ^{2}\beta '}{\operatorname {d} \alpha '^{2}}}}}{{\frac {1+\left({\frac {\operatorname {d} \beta }{\operatorname {d} \alpha }}\right)^{2}}{\frac {\operatorname {d} ^{2}\beta }{\operatorname {d} \alpha ^{2}}}}{\frac {\operatorname {d} \beta }{\operatorname {d} \alpha }}-{\frac {1+\left({\frac {\operatorname {d} \beta '}{\operatorname {d} \alpha '}}\right)^{2}}{\frac {\operatorname {d} ^{2}\beta '}{\operatorname {d} \alpha '^{2}}}}{\frac {\operatorname {d} \beta '}{\operatorname {d} \alpha '}}}}.$

On trouve d’après cela

$1+\operatorname {Tang} r.m\operatorname {Tang} .n=0\,;$