![{\displaystyle 2a=\mathrm {CG+CH} =k\left\{{\frac {\operatorname {Sin} .\alpha }{\operatorname {Sin} .(\beta +\alpha )}}+{\frac {\operatorname {Sin} .\alpha }{\operatorname {Sin} .(\beta -\alpha )}}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85b6e3b5a763283926c7f2a87b9a4c07a4cb70d0)
![{\displaystyle =2k{\frac {\operatorname {Sin} .\alpha \operatorname {Cos} .\alpha \operatorname {Sin} .\beta }{\operatorname {Sin} .(\beta +\alpha )\operatorname {Sin} .(\beta -\alpha )}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5abc079153c48ffcb52e9230cdff1c935515668)
d’où
![{\displaystyle a=k.{\frac {\operatorname {Sin} .\alpha \operatorname {Cos} .\alpha \operatorname {Sin} .\beta }{\operatorname {Sin} .(\beta +\alpha )\operatorname {Sin} .(\beta -\alpha )}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29126067fffbe08501dfbba0e646e8cd7ad4395c)
En outre, en représentant par
le second axe, on devra avoir
![{\displaystyle \mathrm {\frac {{\overline {CD}}^{2}}{CG.CH}} ={\frac {b^{2}}{a^{2}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3698845e57fc6a1fbfa30d7a83c0786eb6baba99)
d’où
![{\displaystyle b^{2}=a^{2}.\mathrm {\frac {{\overline {CD}}^{2}}{CG.CH}} \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0826937b96bc0b3da416660caed5e45053d6796)
substituant donc, et extrayant la racine quarrée, on trouvera
![{\displaystyle b=k.{\frac {\operatorname {Sin} .\alpha \operatorname {Sin} .\beta }{\sqrt {\operatorname {Sin} .(\beta +\alpha )\operatorname {Sin} .(\beta -\alpha )}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bce08f373f50c3c1b9ab2ef477b512562ee2b75)
Si ensuite on désigne le paramètre par
on aura, comme l’on sait,
![{\displaystyle p={\frac {2b^{2}}{a}}=2k\operatorname {Tang} .\alpha \operatorname {Sin} .\beta .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbd2a6e65f7b942fc890c0b4e027b665349ffced)
Désignant encore par
l’excentricité, on aura
![{\displaystyle e^{2}=a^{2}-b^{2}=k^{2}.{\frac {\operatorname {Sin} .^{2}\alpha \operatorname {Sin} .^{2}\beta \operatorname {Cos} .^{2}\beta }{\operatorname {Sin} .^{2}(\beta +\alpha )\operatorname {Sin} .^{2}(\beta -\alpha )}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75cb4e0be24b320c575217c9bb29ec8a87c1a910)
d’où
![{\displaystyle e=k.{\frac {\operatorname {Sin} .\alpha \operatorname {Sin} .\beta \operatorname {Cos} .\beta }{\operatorname {Sin} .(\beta +\alpha )\operatorname {Sin} .(\beta -\alpha )}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f62880783d3497cb4ec4e240010ec90e555371c)
d’après quoi le rapport de l’excentricité au demi-grand axe sera
![{\displaystyle {\frac {e}{a}}={\frac {\operatorname {Cos} .\beta }{\operatorname {Cos} .\alpha }}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0767330d12f30e71d16088eade3a13a8e4db794e)
Si l’on désigne par
la distance d’un sommet au foyer le plus voisin, on aura
![{\displaystyle c=a-e=k.{\frac {\operatorname {Sin} .\alpha \operatorname {Sin} .\beta (\operatorname {Cos} .\alpha -\operatorname {Cos} .\beta )}{\operatorname {Sin} .(\beta +\alpha )\operatorname {Sin} .(\beta -\alpha )}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b8fb45af42d32ae06fe30f0ae579d12c8e9ac12)
ou encore