The locus of the centroid of the triangle formed

Solution:

Given hyperbola is

$16(x+1)^{2}-9(y-2)^{2}=164+16-36=144$

$\Rightarrow \frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1$

Eccentricity, $\mathrm{e}=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$

$\Rightarrow$ foci are $(4,2)$ and $(-6,2)$

Let the centroid be $(\mathrm{h}, \mathrm{k})$

$\& \mathrm{~A}(\alpha, \beta)$ be point on hyperbola

So $\mathrm{h}=\frac{\alpha-6+4}{3}, \mathrm{k}=\frac{\beta+2+2}{3}$

$\Rightarrow \alpha=3 \mathrm{~h}+2, \beta=3 \mathrm{k}-4$

$(\alpha, \beta)$ lies on hyperbola so

$16(3 \mathrm{~h}+2+1)^{2}-9(3 \mathrm{k}-4-2)^{2}=144$

$\Rightarrow 144(\mathrm{~h}+1)^{2}-81(\mathrm{k}-2)^{2}=144$

$\Rightarrow 16\left(\mathrm{~h}^{2}+2 \mathrm{~h}+1\right)-9\left(\mathrm{k}^{2}-4 \mathrm{k}+4\right)=16$

$\Rightarrow 16 \mathrm{x}^{2}-9 \mathrm{y}^{2}+32 \mathrm{x}+36 \mathrm{y}-36=0$