x^2+2y^2+z^2+2kxz=1表示一单叶双曲面,则k满足条件

x^2 + 2y^2 + z^2 + 2kxz = x^2 + 2kxz + (kz)^2 + 2y^2 + (1-k^2)z^2= (x+kz)^2 + 2y^2 + (1-k^2)z^2 = 1双曲面,|k| > 1.单叶,.奇怪.|k| > 1时,(x+kz)^2 + 2y^2 + (1-k^2)z^2 = 1 总是双叶的双曲面啊.难道要 |k| = ...