Hyperbola

:-(

The equation of the asymptotes will differ from that of the hyperbola by a constant and the conjugate hyperbola differs from that of the asymptotes by the same constant.
So writing down the equation of the asymptotes as
xy-3y-4x+7+λ=0
determine λ using the fact that the above equation is a pair of straight lines, and hence the equation of the conjugate hyperbola will be
xy-3y-4x+7+2λ=0.

Note: The asymptotes of the hyperbola given by the general equation of the second degree
ax² + 2hxy + by² + 2gx + 2fy +c = 0
is given by
ax^2+2hxy+by^2+2gx+2fy+c+\dfrac{\Delta }{h^2-ab}=0
and the equation of the conjugate hyperbola is
ax^2+2hxy+by^2+2gx+2fy+c+2\dfrac{\Delta }{h^2-ab}=0

no...you evaluate λ using the fact that xy-3y-4x+7+λ=0 represents a pair of straight lines.

Got the answer.
Thank you :-)