This problem asks us to find the locus of the centroid of a triangle formed by a variable point on a hyperbola and its two fixed foci. This involves understanding the centroid formula, standardizing the hyperbola equation, finding its foci, and then using substitution to derive the locus.

1. Key Concept: The Centroid of a Triangle

The centroid of a triangle is the point where its three medians intersect. If the vertices of a triangle are $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$, then the coordinates of its centroid $G(h, k)$ are given by the formula: $G(h, k) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$ In this problem, the three vertices are a generic point $P(x_P, y_P)$ on the hyperbola, and the two foci, $F_1$ and $F_2$. We will use this formula to establish a relationship between the coordinates of $P$ and the coordinates of the centroid $G$.

2. Standardizing the Hyperbola Equation and Finding its Foci

The given equation of the hyperbola is: $16{x^2} - 9{y^2} + 32x + 36y - 164 = 0$ Our goal is to transform this general form into the standard form of a hyperbola, which will allow us to easily identify its center, $a$, $b$, and subsequently its foci. The standard form for a hyperbola with a horizontal transverse axis is $\frac{(x-h_c)^2}{a^2} - \frac{(y-k_c)^2}{b^2} = 1$, and for a vertical transverse axis is $\frac{(y-k_c)^2}{a^2} - \frac{(x-h_c)^2}{b^2} = 1$.

Step 1: Group terms involving $x$ and $y$, and move the constant term to the right-hand side. $(16x^2 + 32x) - (9y^2 - 36y) = 164$

• Explanation: We group the $x$ terms together and the $y$ terms together to prepare for completing the square. It's crucial to handle the negative sign correctly for the $y$ terms: $-(9y^2 - 36y)$ is equivalent to $-9y^2 + 36y$. A common mistake is to write $-(9y^2 + 36y)$, which changes the sign of the $36y$ term.

Step 2: Factor out the coefficients of $x^2$ and $y^2$. $16(x^2 + 2x) - 9(y^2 - 4y) = 164$

• Explanation: Factoring out the leading coefficients inside the parentheses simplifies the expressions, making it easier to apply the completing the square technique. The coefficients must be factored out before completing the square.

Step 3: Complete the square for both the $x$ and $y$ expressions.

• For the $x$-terms: To complete the square for $x^2 + 2x$, we add $(\frac{2}{2})^2 = 1^2 = 1$. Since this $(x^2 + 2x + 1)$ expression is multiplied by $16$, we are effectively adding $16 \times 1 = 16$ to the left side of the equation.
• For the $y$-terms: To complete the square for $y^2 - 4y$, we add $(\frac{-4}{2})^2 = (-2)^2 = 4$. Since this $(y^2 - 4y + 4)$ expression is multiplied by $-9$, we are effectively adding $-9 \times 4 = -36$ to the left side of the equation.
• To maintain the equality, we must add these same effective amounts to the right side of the equation.

$16(x^2 + 2x + 1) - 9(y^2 - 4y + 4) = 164 + 16 - 36$ $16(x+1)^2 - 9(y-2)^2 = 144$

• Explanation: Completing the square transforms quadratic expressions into perfect square trinomials, which can then be written in the form $(x-h_c)^2$ or $(y-k_c)^2$. This step directly reveals the coordinates of the center of the hyperbola.

Step 4: Divide the entire equation by the constant on the right-hand side to get the standard form. $\frac{16(x+1)^2}{144} - \frac{9(y-2)^2}{144} = \frac{144}{144}$ $\frac{(x+1)^2}{9} - \frac{(y-2)^2}{16} = 1$

• Explanation: The standard form requires the right-hand side to be $1$. Dividing by $144$ achieves this and allows us to directly read off $a^2$ and $b^2$.

From this standard form, we can identify the following characteristics of the hyperbola:

• The center of the hyperbola $C(h_c, k_c) = (-1, 2)$.
• $a^2 = 9 \implies a = 3$. This is the distance from the center to each vertex along the transverse axis.
• $b^2 = 16 \implies b = 4$. This is the distance from the center to each co-vertex along the conjugate axis.
• Since the term with $(x+1)^2$ is positive, the transverse axis is parallel to the x-axis.

Step 5: Calculate the eccentricity and locate the foci. For a hyperbola, the relationship between $a$, $b$, and the eccentricity $e$ is given by $c^2 = a^2 + b^2$, where $c = ae$. So, $a^2e^2 = a^2 + b^2$, which implies $e^2 = 1 + \frac{b^2}{a^2}$. $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{9+16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$ The distance from the center to each focus is $ae$. $ae = 3 \times \frac{5}{3} = 5$ Since the transverse axis is parallel to the x-axis, the foci are located at $(h_c \pm ae, k_c)$. Therefore, the foci are: $F_1 = (-1 - 5, 2) = (-6, 2)$ $F_2 = (-1 + 5, 2) = (4, 2)$

3. Setting up the Centroid Equations

Let $P(x_P, y_P)$ be any point on the hyperbola. Let $G(h, k)$ be the centroid of the triangle $PF_1F_2$. The vertices of the triangle are $P(x_P, y_P)$, $F_1(-6, 2)$, and $F_2(4, 2)$.

Using the centroid formula: For the x-coordinate of the centroid: $h = \frac{x_P + (-6) + 4}{3} = \frac{x_P - 2}{3}$ For the y-coordinate of the centroid: $k = \frac{y_P + 2 + 2}{3} = \frac{y_P + 4}{3}$

• Explanation: We have now expressed the coordinates of the centroid $(h, k)$ in terms of the coordinates of the point $P(x_P, y_P)$ on the hyperbola. To find the locus of the centroid, we need an equation solely in terms of $h$ and $k$. We can achieve this by expressing $x_P$ and $y_P$ in terms of $h$ and $k$, and then substituting these into the hyperbola's equation.

From the centroid equations, we can express $x_P$ and $y_P$: $x_P - 2 = 3h \implies x_P = 3h + 2$ $y_P + 4 = 3k \implies y_P = 3k - 4$

4. Substituting into the Hyperbola Equation to Find the Locus

Since $P(x_P, y_P)$ lies on the hyperbola $\frac{(x_P+1)^2}{9} - \frac{(y_P-2)^2}{16} = 1$, we can substitute the expressions for $x_P$ and $y_P$ (in terms of $h$ and $k$) into this equation:

$\frac{((3h+2)+1)^2}{9} - \frac{((3k-4)-2)^2}{16} = 1$ $\frac{(3h+3)^2}{9} - \frac{(3k-6)^2}{16} = 1$ Factor out $3$ from the terms in the numerators: $\frac{(3(h+1))^2}{9} - \frac{(3(k-2))^2}{16} = 1$ $\frac{9(h+1)^2}{9} - \frac{9(k-2)^2}{16} = 1$ $(h+1)^2 - \frac{9(k-2)^2}{16} = 1$ To eliminate the fraction, multiply the entire equation by $16$: $16(h+1)^2 - 9(k-2)^2 = 16$ Now, expand the squared terms: $16(h^2 + 2h + 1) - 9(k^2 - 4k + 4) = 16$ $16h^2 + 32h + 16 - 9k^2 + 36k - 36 = 16$ Rearrange the terms and combine constants: $16h^2 - 9k^2 + 32h + 36k + 16 - 36 - 16 = 0$ $16h^2 - 9k^2 + 32h + 36k - 36 = 0$ Finally, to represent the locus, we replace $h$ with $x$ and $k$ with $y$: $16x^2 - 9y^2 + 32x + 36y - 36 = 0$

• Explanation: This final equation describes the relationship between the $x$ and $y$ coordinates of the centroid $G$. This is the locus we were looking for. Notice that this new equation is also a hyperbola, which is expected since the centroid's coordinates are linearly related to the hyperbola's coordinates.

5. Final Check and Conclusion

Comparing our derived locus equation with the given options: $16x^2 - 9y^2 + 32x + 36y - 36 = 0$ This matches option (A).

Key Takeaway: This problem demonstrates a common technique for finding the locus of a point related to a curve. The general strategy is:

1. Define the coordinates of the point whose locus is sought (e.g., $G(h,k)$).
2. Establish relationships between $(h,k)$ and the coordinates of the variable point on the given curve (e.g., $P(x_P,y_P)$).
3. Express the coordinates of the variable point in terms of $(h,k)$.
4. Substitute these expressions into the equation of the given curve.
5. Simplify the resulting equation and replace $h$ with $x$ and $k$ with $y$ to get the locus equation.

This approach is applicable to finding loci for various geometric constructions involving points on curves.