AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

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