若方程x^2+(k-1)y^2-3ky+2k=0 表示双曲线,则实数k的取值范围?

需要配方.
0=x^2+(k-1)y^2-3ky+2k=x^2-(1-k)[y^2+3ky/(1-k)]+2k
=x^2-(1-k){[y+3/2*k/(1-k)]^2+(1-k)[3/2*k/(1-k)]^2+2k
=x^2-(1-k){[y+3/2*k/(1-k)]^2+k(k+8)/[4(1-k)]
故有
x^2-(1-k){[y+3/2*k/(1-k)]^2=k(k+8)/[4(k-1)]
要使为双曲线,必须
1-k>0,且k(k+8)/[4(k-1)]≠0
于是得k