Let ax+by=1 be the chord ...(i) Making the equation of hyperbola homogeneous using Eq. (i), we get 3x2−y2+(−2x+4y)(ax+by)=0 or (3−2a)x2+(−1+4b)y2+(−2b+4a)xy=0 Since, the angle subtended at the origin is a right angle ∴ Coefficient of x2+ Coefficient of y2=0 ⇒(3−2a)+(−1+4b)=0 ⇒a=2b+1 ∴ Chords are (2b+1)x+by−1=0 or b(2+y)+(x−1)=0 which clearly pass through the fixed point (1,-2)