Correct Answer - B
Let the asymptotes be `2x+3y+lambda_(1)=0 and x+2y+lambda_(2)=0`.
It will pass through the centre (1, 2). Hence,
`lambda_(2)=-8, lambda_(2)=-5`
The equation of the hyperbola is
`(2x+3y-8)(x+2y-5)+lambda=0`
It passes through (2, 4). Therefore,
`(4+12-8)(2+8-5)+lambda=0 or lambda=-40`
Hence, equation of hyperbola is
`(2x+3y-8)(x+2y-5)=40`