The locus of the centroid of the triangle formed by any point P on the hyperbola 16x^2-9y^2+32x+36y-164=0 and its foci is

Given hyperbola is

$$16{x}^{2}-9{y}^{2}+32x+36y-164=0$$16x2−9y2+32x+36y−164=0

$$\Rightarrow 16{(x+1)}^{2}-9{(y-2)}^{2}=164+16-36$$⇒16(x+1)2−9(y−2)2=164+16−36

$$\Rightarrow 16{(x+1)}^{2}-9{(y-2)}^{2}=144....(1)$$⇒16(x+1)2−9(y−2)2=144....(1)

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

Compare the above equation with $\Rightarrow \frac{(x-h{)}^{2}}{{a}^{2}}-\frac{(y-k{)}^{2}}{{b}^{2}}=1$⇒a2(x−h)2​−b2(y−k)2​=1

We get, $h=-1,k=2,{a}^{2}=9$h=−1,k=2,a2=9 and ${b}^{2}=16$b2=16

$\therefore$∴ Eccentricity, $e=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$e=1+916​​=35​

We know that, focii are $(h+ae,k)$(h+ae,k) and $(h-ae,k)$(h−ae,k)

Hence, focii are $(4,2)$(4,2) and $(-6,2)$(−6,2)

Let the centroid be $(h,k)$(h,k) & $A(\alpha ,\beta )$A(α,β) be point on hyperbola

So $h=\frac{\alpha -6+4}{3},k=\frac{\beta +2+2}{3}$h=3α−6+4​,k=3β+2+2​

$$\Rightarrow \alpha =3h+2,\beta =3k-4$$⇒α=3h+2,β=3k−4

Put the values of $\alpha ,\beta$α,β in the equation$(1)$(1),

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

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

$$\Rightarrow 16({h}^{2}+2h+1)-9({k}^{2}-4k+4)=16$$⇒16(h2+2h+1)−9(k2−4k+4)=16

$$\Rightarrow 16{h}^{2}-9{k}^{2}+32h+36k-36=0$$⇒16h2−9k2+32h+36k−36=0

So, the locus of the centroid is

$$16{x}^{2}-9{y}^{2}+32x+36y-36=0$$16x2−9y2+32x+36y−36=0