Given hyperbola is
16x2−9y2+32x+36y−164=0
⇒16(x+1)2−9(y−2)2=164+16−36
⇒16(x+1)2−9(y−2)2=144....(1)
⇒9(x+1)2−16(y−2)2=1
Compare the above equation with ⇒a2(x−h)2−b2(y−k)2=1
We get, h=−1,k=2,a2=9 and b2=16
∴ Eccentricity, e=1+916=35
We know that, focii are (h+ae,k) and (h−ae,k)
Hence, focii are (4,2) and (−6,2)

Let the centroid be (h,k) & A(α,β) be point on hyperbola
So h=3α−6+4,k=3β+2+2
⇒α=3h+2,β=3k−4
Put the values of α,β in the equation(1),
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
⇒16h2−9k2+32h+36k−36=0
So, the locus of the centroid is
16x2−9y2+32x+36y−36=0