199
D’UNE ELLIPSE.
On déduit de là, après les réductions
![{\displaystyle {\begin{aligned}\operatorname {Tang} .(P-\phi )&={\tfrac {p(r-q)+q(p-r)\operatorname {Cos} .(Q-P)+r(q-p).\operatorname {Cos} .(R-P)}{q(p-r)\operatorname {Sin} .(Q-P)+r(q-p)\operatorname {Sin} .(R-P)}},\\\operatorname {Tang} .(Q-\phi )&={\tfrac {p(r-q)\operatorname {Cos} .(P-Q)+q(p-r)+r(q-p)\operatorname {Cos} .(R-Q)}{p(r-q)\operatorname {Sin} .(P-Q)+r(q-p)\operatorname {Sin} .(R-Q)}},\\\operatorname {Tang} .(R-\phi )&={\tfrac {p(r-q)\operatorname {Cos} .(P-R)+q(p-r)\operatorname {Cos} .(Q-R)+r(q-p)}{p(r-q)\operatorname {Sin} .(P-R)+q(p-r)\operatorname {Sin} .(Q-R)}}.\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ea7c6bed3914cbab5862935056e74694f2ac0fc)
La nature du problème exige que des tangentes on passe aux cosinus. On y parvient moyennant une certaine fonction, qu’en attendant nous représenterons par
et dont la valeur, que nous nous réservons de simplifier plus loin, peut être exprimée ainsi qu’il suit :
![{\displaystyle F^{2}=\left\{{\begin{aligned}(r-q)^{2}p^{2}&+2rq(p-r)(q-p)\operatorname {Cos} .(R-Q)\\+(p-q)^{2}r^{2}&+2pr(q-p)(r-q)\operatorname {Cos} .(P-R)\\+(q-p)^{2}r^{2}&+2qp(r-q)(p-r)\operatorname {Cos} .(Q-P).\\\end{aligned}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e189f10d50a15e93916bd579362ccdf5e0014fd)
On trouve alors
![{\displaystyle \operatorname {Sin} .\lambda ={\frac {F}{pq\operatorname {Sin} .(Q-P)+qr\operatorname {Sin} .(R-Q)+rp\operatorname {Sin} .(P-R)}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55ad31c465f5883db935fce0feff006d15a4a8c4)
et ensuite
![{\displaystyle {\begin{aligned}F\operatorname {Cos} .(P-\phi )&=q(p-r)\operatorname {Sin} .(Q-P)+r(q-p)\operatorname {Sin} .(R-P),\\F\operatorname {Cos} .(Q-\phi )&=r(q-p)\operatorname {Sin} .(R-Q)+p(r-q)\operatorname {Sin} .(P-Q),\\F\operatorname {Cos} .(R-\phi )&=p(r-q)\operatorname {Sin} .(P-R)+q(p-r)\operatorname {Sin} .(Q-R)\,;\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d9f18c6e4be3826b89c034dac287c18b55a6fd8)
d’où encore
![{\displaystyle {\begin{aligned}\operatorname {Sin} .\lambda \operatorname {Cos} .(P-\phi )&={\tfrac {q(p-r)\operatorname {Sin} .(Q-P)+r(q-p)\operatorname {Sin} .(R-P)}{pq\operatorname {Sin} .(Q-P)+qr\operatorname {Sin} .(R-Q)+rp\operatorname {Sin} .(P-R)}},\\\operatorname {Sin} .\lambda \operatorname {Cos} .(Q-\phi )&={\tfrac {r(q-p)\operatorname {Sin} .(R-Q)+p(r-q)\operatorname {Sin} .(P-Q)}{pq\operatorname {Sin} .(Q-P)+qr\operatorname {Sin} .(R-Q)+rp\operatorname {Sin} .(P-R)}},\\\operatorname {Sin} .\lambda \operatorname {Cos} .(R-\phi )&={\tfrac {p(r-q)\operatorname {Sin} .(P-R)+q(p-r)\operatorname {Sin} .(Q-R)}{pq\operatorname {Sin} .(Q-P)+qr\operatorname {Sin} .(R-Q)+rp\operatorname {Sin} .(P-R)}}.\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e9b2579aac40594f7e96f1201c5db83e66aa70a)
De là résulte l’égalité suivante