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

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

$16 x^{2} - 9 y^{2} + 32 x + 36 y - 144 = 0$

$9 x^{2} - 16 y^{2} + 36 x + 32 y - 144 = 0$

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

answer is A.

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Detailed Solution

Given hyperbola is $16 x^{2} + 32 x - 9 y^{2} + 36 y = 164$
$\begin{array}{l} \Rightarrow 16 {(x + 1)}^{2} - 9 {(y - 2)}^{2} = 164 + 16 - 36 = 144 \\ \Rightarrow \frac{{(x + 1)}^{2}}{9} - \frac{{(y - 2)}^{2}}{16} = 1 \end{array}$

Eccentricity, $e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{16}{9}} = \frac{5}{3}$
$\begin{array}{l} \Rightarrow f o c i are (- 1 \pm a e, 0 + 2) \\ \Rightarrow (4, 2) and (- 6, 2) \end{array}$

Let the centroid of $Δ ABC$
be $(h, k) & A (α, β)$ be a point on hyperbola
$\begin{array}{l} ∴ h = \frac{α - 6 + 4}{3}, k = \frac{β + 2 + 2}{3} \\ \Rightarrow α = 3 h + 2, β = 3 k - 4 \end{array}$

Since $(α, β)$ lies on hyperbola
$\begin{array}{l} ∴ 16 {(3 h + 2 + 1)}^{2} - 9 {(3 k - 4 - 2)}^{2} = 144 \\ \Rightarrow 144 {(h + 1)}^{2} - 81 {(k - 2)}^{2} = 144 \\ \Rightarrow 16 (h^{2} + 2 h + 1) - 9 (k^{2} - 4 k + 4) = 16 \\ \Rightarrow 16 x^{2} - 9 y^{2} + 32 x + 36 y - 36 = 0 \end{array}$
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