chalte chalte conjugate hyperbola..

The equation of the asymptotes will differ from the conjugate hyperbola by a constant. So the equation of the asymptotes is of the form
xy-2x-4y+k=0
Since this equation represents a pair of straight line, the discriminant must be zero. So
2(-2)(-1)(1/2)-k(1/2)² = 0
This gives k = 8.
Hence the asymptotes are xy-2x-4y+8=0
Since the conjugate hyperbola differs from the asymptotes by the same constant as the asymptote does from the hyperbola, the equation of the conjugate hyperbola becomes
xy - 2x - 4y +12=0

thanks sir [1]