If in the form
x 2 a 2 + y 2 b 2 − z 2 c 2 = 1 {x^2 \over a^2}+{y^2\over b^2}-{z^2 \over c^2}=1 a 2 x 2 + b 2 y 2 − c 2 z 2 = 1 then let z = c v , v z=cv, v z = c v , v as one parameter, and then, the equation can be rewritten in the following way:
x 2 a 2 + y 2 b 2 − c 2 v 2 c 2 = 1 {x^2 \over a^2}+{y^2\over b^2}-{c^2v^2 \over c^2}=1 a 2 x 2 + b 2 y 2 − c 2 c 2 v 2 = 1 x 2 a 2 + y 2 b 2 = 1 + v 2 {x^2 \over a^2}+{y^2\over b^2}=1+v^2 a 2 x 2 + b 2 y 2 = 1 + v 2 Let x = a 1 + v 2 cos u , y = b 1 + v 2 sin u . x=a\sqrt{1+v^2}\cos u, y=b\sqrt{1+v^2}\sin u. x = a 1 + v 2 cos u , y = b 1 + v 2 sin u . Then
( a 1 + v 2 cos u ) 2 a 2 + ( b 1 + v 2 sin u ) 2 b 2 = 1 + v 2 , {(a\sqrt{1+v^2}\cos u)^2 \over a^2}+{(b\sqrt{1+v^2}\sin u)^2\over b^2}=1+v^2, a 2 ( a 1 + v 2 cos u ) 2 + b 2 ( b 1 + v 2 sin u ) 2 = 1 + v 2 ,
x = a 1 + v 2 cos u , x=a\sqrt{1+v^2}\cos u, x = a 1 + v 2 cos u , y = b 1 + v 2 sin u , y=b\sqrt{1+v^2}\sin u, y = b 1 + v 2 sin u , z = c v , z=cv, z = c v , u ∈ [ 0 , 2 π ) , v ∈ [ − 1 c , 1 c ] . u\in[0, 2\pi), v\in\big [-{ 1\over c} , {1 \over c} \big]. u ∈ [ 0 , 2 π ) , v ∈ [ − c 1 , c 1 ] . Other parameterizations include
x = a cosh v cos u , x=a\cosh{v}\cos{u}, x = a cosh v cos u ,
y = b cosh v sin u , y=b\cosh{v}\sin{u}, y = b cosh v sin u ,
z = c sinh v , z=c\sinh{v}, z = c sinh v ,
u ∈ [ 0 , 2 π ) , v ∈ [ − sinh − 1 ( 1 c ) , sinh − 1 ( 1 c ) ] . u\in[0, 2\pi), v\in\big [-\sinh^{-1}({ 1\over c}) , \sinh^{-1}({ 1\over c})]. u ∈ [ 0 , 2 π ) , v ∈ [ − sinh − 1 ( c 1 ) , sinh − 1 ( c 1 )] .
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