Official solution
Given hyperbola is
\[16{(x + 1)^2} - 9{(y - 2)^2} = 164 + 16 - 36 = 144\]
\[ \Rightarrow {{{{(x + 1)}^2}} \over 9} - {{{{(y - 2)}^2}} \over {16}} = 1\]
Eccentricity, \[e = \sqrt {1 + {{16} \over 9}} = {5 \over 3}\]
\[\Rightarrow\] foci are (4, 2) and (-6, 2)
Let the centroid be (h, k) \& A(\[\alpha\], \[\beta\]) be point on hyperbola.
So, \[h = {{\alpha - 6 + 4} \over 3},k = {{\beta + 2 + 2} \over 3}\]
\[ \Rightarrow \alpha = 3h + 2,\beta = 3k - 4\]
(\[\alpha\], \[\beta\]) lies on hyperbola so
\[16{(3h + 2 + 1)^2} - 9{(3k - 4 - 2)^2} = 144\]
\[ \Rightarrow 144{(h + 1)^2} - 81{(k - 2)^2} = 144\]
\[ \Rightarrow 16({h^2} + 2h + 1) - 9({k^2} - 4k + 4) = 16\]
\[ \Rightarrow 16{x^2} - 9{y^2} + 32x + 36y - 36 = 0\]