Given, hyperbola is 16x2−9y2+32x−36y−164=0 ⇒16(x+1)2−9(y−2)2 =164+16−36=144
⇒
(x+1)2
9
−
(4−2)2
16
=1 Eccentricity, e=√1+
16
9
=
5
3
⇒ Foci are (4,2) and (−6,2).
Let the centroid be (h,k) and A(α,β) be point on hyperbol(a) So, h=
α−6+4
3
,k=
β+2+2
3
⇒α=3h+2,β=2k−4 (α,β) lies on hyperbola, so 16(3h+2+1)2−9(3k−4−2)2=144 ⇒144(h+1)2−81(k−2)2=144 ⇒16(h2+2h+1)−9(k2−4k+4)=16 ⇒16x2−9y2+32x+36y−36=0