the transverse of the conic:3x^2-y^2+2y-1=0 is ???
in standard form, your hyperbola is
3x^2-y^2+2y-1=0
3x^2 - (y-1)^2 = 0
x^2 - (y-1)^2/3 = 0
Your hyperbola is just two intersecting lines, so its transverse is zero! I suspect a typo.
3x^2-y^2+2y-1=0
3x^2 - (y-1)^2 = 0
x^2 - (y-1)^2/3 = 0
Your hyperbola is just two intersecting lines, so its transverse is zero! I suspect a typo.