Let the points are, A(2,0),A′(−2,0) and S(−3,0) ⇒ Centre of hyperbola is O(0,0) AA′=2a⇒4=2a⇒a=2 ∵ Distance between the centre and foci is ae. ∴OS=ae⇒3=2e⇒e=
3
2
⇒b2=a2(e2−1)=a2e2−a2=9−4=5 ⇒ Equation of hyperbola is
x2
4
−
y2
5
=1 . . . . (i) ∵(6,552) does not satisfy eq (i). ∠(6,5√2) does not lie on this hyperbola.