374
QUESTIONS
Pour deuxième exemple, prenons l’hyperbole équilatère ayant
pour équation
![{\displaystyle 2xy-a^{2}=Z=0\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/821fe32610ac4c641121a5c41999b2e30af18565)
qui donne
![{\displaystyle X=\left({\frac {\operatorname {d} Z}{\operatorname {d} x}}\right)=2y,\qquad Y=\left({\frac {\operatorname {d} Z}{\operatorname {d} y}}\right)=2x.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/154bc8499c18ba254a09ca58f7f30f38f9a8b6d5)
Les équations du problème seront
![{\displaystyle {\begin{array}{cr}(x'-x'')^{2}+(y'-y'')^{2}=4c^{2},&{\text{(I)}}\\\\2x'y'=a^{2},\qquad {\text{(II)}}\qquad 2x''y''=a^{2}&{\text{(III)}}\\\\(y'-y'')x-(x'-x'')y+(x'y''-y'x'')=0,&{\text{(IV)}}\\\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2147491ddf8251b33d0a567506e742e2dc6ccc00)
![{\displaystyle (x'y''-x''y')\left[(x'-x'')x+(y'-y'')y\right]=(x'y''+x''y')\left[(x'-x'')^{2}-(y'-y'')^{2}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fde83c56835662191851f1b6e257be81af45ccbd)
(V)
Si, au moyen des équations (II, III), on chasse
des trois autres, elles deviendront,
![{\displaystyle {\begin{array}{cr}(x'-x'')^{2}\left(4x'^{2}x''^{2}+a^{4}\right)=16c^{2}x'^{2}x''^{2},&{\text{(VI)}}\\\\2yx'x''+a^{2}x=a^{2}(x'+x''),&{\text{(VII)}}\\\\2x'x''(x'+x'')(2xx'x''-a^{2}y)=(x'^{2}+x''^{2})(4x'^{2}x''-a^{4}).&{\text{(VIII)}}\\\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff08ef9128b083f1efeb10123eda397307025682)
Posant alors
![{\displaystyle x'+x''=t,\qquad x'y'=u,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc8a0a5189fa8507121168258ca786ce88f78749)
d’où
![{\displaystyle x'^{2}+x''^{2}=t^{2}-2u,\qquad (x'-x'')^{2}=t^{2}-4u,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0baa01d9c8f2fc90c9ea1a0a102fb22570f10ce1)
il viendra, en substituant
![{\displaystyle (t^{2}-4u)(4u^{2}+a^{4})=16c^{2}u^{2},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fc64bb57181362d8248a3d2278f6405d7b7ff46)