Given hyperbola is ‌9x2−16y2−36x+96y−252=0 ‌⇒9(x2−4x)−16(y2−6y)=252 ‌⇒9(x2−4x+4)−16(y2−6y+9) ‌=252+36−144 ‌⇒9(x−2)2−16(y−3)2=144 ‌⇒‌
(x−2)2
16
−‌
(y−3)2
9
=1 . . . (i) Put x=X+2 and y=Y+3 ∴ Equation (i) becomes ‌
X2
16
−‌
Y2
9
=1 Now, vertices are X=±a wehre a=4 and Y=0 Hence, vertices are (6,3),(−2,3).