楕円
ii) − π / 2 < θ 1 < 0 ≤ θ 2 ≤ π / 2 {\displaystyle -\pi /2<\theta _{1}<0\leq \theta _{2}\leq \pi /2} L = b ( E ( v 2 , e ) + E ( v 1 , e ) ) {\displaystyle L=b\,(E(v_{2},e)+E(v_{1},e))} v 1 = t h 2 v ( − θ 1 ) , v 2 = t h 2 v ( θ 2 ) {\displaystyle v_{1}=th2v(-\theta _{1}),\,v_{2}=th2v(\theta _{2})}
iii) − π / 2 ≤ θ 1 ≤ θ 2 < 0 {\displaystyle -\pi /2\leq \theta _{1}\leq \theta _{2}<0} L = b ( E ( v 2 , e ) − E ( v 1 , e ) ) {\displaystyle L=b\,(E(v_{2},e)-E(v_{1},e))} v 1 = t h 2 v ( − θ 2 ) , v 2 = t h 2 v ( − θ 1 ) {\displaystyle v_{1}=th2v(-\theta _{2}),\,v_{2}=th2v(-\theta _{1})}
ここで
t h 2 v ( θ ) = tan − 1  ( ( a / b ) tan  θ ) {\displaystyle th2v(\theta )=\tan ^{-1}((a/b)\tan \theta )}
(ただし、 t h 2 v ( ± π / 2 ) = ± π / 2 {\displaystyle th2v(\pm \pi /2)=\pm \pi /2} とする)
である。