Answer on Question #77265, Math / Geometry
σ(u,v)=(sinhu⋅sinhv,sinhv,sinhu⋅coshv,sinhu)∂u∂σ=(coshu⋅sinhv,0,coshu⋅coshv,coshu)∂v∂σ=(sinhu⋅coshv,coshv,sinhu⋅sinhv,0)E=cosh2u⋅sinh2v+cosh2u⋅cosh2v+cosh2u=2cosh2u⋅cosh2vF=2sinhu⋅coshu⋅sinhv⋅coshvG=sinh2u⋅cosh2v+cosh2v+sinh2u⋅sinh2v
Then, 1(x,y)=Ex2+2Fxy+Gy2
σ(u,v)=(u−v,u+v,u2+v2)∂u∂σ=(1,1,2u)∂v∂σ=(−1,1,2v)E=2+4u2F=4uvG=2+4v2
Then, 1(x,y)=Ex2+2Fxy+Gy2