377
RÉSOLUES.
![{\displaystyle mn-m'n'=\operatorname {Tang} .\alpha \operatorname {Tang} .\beta \operatorname {Tang} .\alpha '\operatorname {Tang} .\beta '}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5126edcbfe9c8c6dff1697d4c7d8125436bb0478)
![{\displaystyle ={\frac {\operatorname {Sin} .\alpha \operatorname {Sin} .\beta \operatorname {Cos} .\alpha '\operatorname {Cos} .\beta '-\operatorname {Cos} .\alpha \operatorname {Cos} .\beta \operatorname {Sin} .\alpha '\operatorname {Sin} .\beta '}{\operatorname {Cos} .\alpha \operatorname {Cos} .\beta \operatorname {Cos} .\alpha '\operatorname {Cos} .\beta '}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/885c391f9159f4b9c8c524821b80000dd2fdbec0)
![{\displaystyle {\begin{array}{cll}m+n&=\operatorname {Tang} .\alpha +\operatorname {Tang} .\beta &={\frac {\operatorname {Sin} .(\alpha +\beta )}{\operatorname {Cos} .\alpha \operatorname {Cos} .\beta }},\\m'+n'&=\operatorname {Tang} .\alpha '+\operatorname {Tang} .\beta '&={\frac {\operatorname {Sin} .(\alpha '+\beta ')}{\operatorname {Cos} .\alpha '\operatorname {Cos} .\beta '}},\\1+mn&=1+\operatorname {Tang} .\alpha \operatorname {Tang} .\beta &={\frac {\operatorname {Cos} .(\alpha -\beta )}{\operatorname {Cos} .\alpha \operatorname {Cos} .\beta }},\\1+m'n'&=1+\operatorname {Tang} .\alpha '\operatorname {Tang} .\beta '&={\frac {\operatorname {Cos} .(\alpha '-\beta ')}{\operatorname {Cos} .\alpha '\operatorname {Cos} .\beta '}}\,;\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4d05f751cc7439f44d8d58999aec85ccaa2cae6)
il viendra donc, en substituant,
![{\displaystyle \operatorname {Tang} .2z={\frac {2\left(\operatorname {Sin} .\alpha \operatorname {Sin} .\beta \operatorname {Cos} .\alpha '\operatorname {Cos} .\beta '-\operatorname {Cos} .\alpha \operatorname {Cos} .\beta \operatorname {Sin} .\alpha '\operatorname {Sin} .\beta '\right)}{\operatorname {Cos} .(\alpha -\beta )\operatorname {Sin} .(\alpha '+\beta ')-\operatorname {Cos} .(\alpha '-\beta ')\operatorname {Sin} .(\alpha +\beta )}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66b0ddbc4cef9f61f91d313ff8d6e2e968fbb973)
formule que l’on pourra encore écrire ainsi
![{\displaystyle \operatorname {Tang} .2z={\frac {\operatorname {Cos} .(\alpha '+\beta ')\operatorname {Cos} .(\alpha -\beta )-\operatorname {Cos} .(\alpha +\beta )\operatorname {Cos} .(\alpha '-\beta ')}{\operatorname {Sin} .(\alpha '+\beta ')\operatorname {Cos} .(\alpha -\beta )-\operatorname {Sin} .(\alpha +\beta )\operatorname {Cos} .(\alpha '-\beta ')}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddfbff80c3ce918f12385a9a5bec7376aee58d3c)
L’angle
, que fait l’un des diamètres principaux avec notre droite indéfinie, étant connu par cette formule, le rapport de grandeur
des deux diamètres principaux sera donné par l’une ou l’autre des équations (1), d’où l’on tire
![{\displaystyle {\frac {y}{x}}={\sqrt {-\operatorname {Tang} .(z+\alpha )\operatorname {Tang} .(z+\beta )}}={\sqrt {-\operatorname {Tang} .(z+\alpha ')\operatorname {Tang} .(z+\beta ')}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bb454a8091f928a46296a48060fad7e77aa544f)
Tout ce qui précède s’applique au surplus littéralement à l’hyperbole, pourvu que, dans les dernières formules, on change le signe
en
sous les radicaux.