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 :

Options

A. `16x^(2)-9y^(2)+32x+36y-36=0`
B. `9x^(2)-16y^(2)+36x+32y-144=0`
C. `16x^(2)-9y^(2)+32x+36y-144=0`
D. `9x^(2)-16y^(2)+36x+32y-36=0`

Answer

To find 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, we will follow these steps:

### Step 1: Rewrite the Hyperbola in Standard Form
We start with the equation of the hyperbola:
\[ 16x^2 - 9y^2 + 32x + 36y - 164 = 0 \]
To convert this into standard form, we will complete the square for the \( x \) and \( y \) terms.

1. Group the \( x \) terms and \( y \) terms:
\[ 16(x^2 + 2x) - 9(y^2 - 4y) - 164 = 0 \]

2. Complete the square:
- For \( x^2 + 2x \), add and subtract \( 1 \):
\[ x^2 + 2x = (x + 1)^2 - 1 \]
- For \( y^2 - 4y \), add and subtract \( 4 \):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

3. Substitute back into the equation:
\[ 16((x + 1)^2 - 1) - 9((y - 2)^2 - 4) - 164 = 0 \]
Simplifying gives:
\[ 16(x + 1)^2 - 9(y - 2)^2 - 16 + 36 - 164 = 0 \]
\[ 16(x + 1)^2 - 9(y - 2)^2 - 144 = 0 \]
\[ 16(x + 1)^2 - 9(y - 2)^2 = 144 \]

4. Divide by 144 to get the standard form:
\[ \frac{(x + 1)^2}{9} - \frac{(y - 2)^2}{16} = 1 \]

### Step 2: Identify Parameters of the Hyperbola
From the standard form, we identify:
- \( a^2 = 9 \) \( \Rightarrow a = 3 \)
- \( b^2 = 16 \) \( \Rightarrow b = 4 \)
- Center \( (h, k) = (-1, 2) \)

### Step 3: Find the Foci of the Hyperbola
The foci of the hyperbola are given by:
\[ c = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
The foci are located at:
\[ S = (h - c, k) = (-1 - 5, 2) = (-6, 2) \]
\[ S' = (h + c, k) = (-1 + 5, 2) = (4, 2) \]

### Step 4: Find the Centroid of Triangle \( PSS' \)
Let \( P \) be any point on the hyperbola, represented as \( (x_1, y_1) \). The centroid \( G \) of triangle \( PSS' \) is given by:
\[ G = \left( \frac{x_1 + (-6) + 4}{3}, \frac{y_1 + 2 + 2}{3} \right) = \left( \frac{x_1 - 2}{3}, \frac{y_1 + 4}{3} \right) \]

### Step 5: Express \( x_1 \) and \( y_1 \) in Terms of \( G \)
Let \( G = (h, k) \). Then:
\[ x_1 = 3h + 2 \]
\[ y_1 = 3k - 4 \]

### Step 6: Substitute into the Hyperbola Equation
Substituting \( x_1 \) and \( y_1 \) into the hyperbola equation:
\[ 16(3h + 2)^2 - 9(3k - 4)^2 + 32(3h + 2) + 36(3k - 4) - 164 = 0 \]

### Step 7: Simplify the Equation
1. Expand and simplify:
- \( 16(9h^2 + 12h + 4) - 9(9k^2 - 24k + 16) + 96h + 64 + 108k - 144 - 164 = 0 \)
- Combine like terms to form a quadratic in \( h \) and \( k \).

2. After simplification, we will arrive at the final locus equation:
\[ 144h^2 - 81k^2 + 288h + 324k - 324 = 0 \]

### Final Answer
The locus of the centroid of the triangle formed by any point \( P \) on the hyperbola and its foci is:
\[ 144x^2 - 81y^2 + 288x + 324y - 324 = 0 \]