The equation of asymptotes is given as, 3x+4y−2=0 2x+y+1=0 The above equation is given by, (3x+4y−2)(2x+y+1)=0 The equation of hyperbola with respect to the above equation is given as, (3x+4y−2)(2x+y+1)=λ Here, the constant is given as λ. The point (1,1) lies on the curve. ∴(3+4−2)(2+1+1)=λ 5(4)=λ=20 The equation of the hyperbola becomes, (3x+4y−2)(2x+y+1)=20 6x2+4y2+11xy−x+2y−22=0 6x2+11xy+4y2−x+2y−22=0