EAMCET Engineering 2010 Solved Paper

Since, the equation of the hyperbola differs from that of the joint equations of the tangents by a constant, therefore the equation of the hyperbola will be of the form
(4x+3y−7)(x−2y−1)+k=0 . . . (i)
∴ Since, the hyperbola passes through the point (2,3)
⇒(8+9−7)(2−6−1)+k=0
⇒(10)(−5)+k=0,−50+k=0
⇒k=50
Hence, from Eq. (i)
(4x+3y−7)(x−2y−1)+50=0
4x2+3xy−7x−8xy−6y2+14y−4x
−3y+7+50=0
4x2−5xy−6y2−11x+11y+57=0