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The equation given is:

\[ 9x^2 - 4y^2 - 126x + 48y - 279 = 0 \]

To determine the type of conic section, we will complete the square for both \(x\) and \(y\).

### Step 1: Group \(x\) and \(y\) terms

\[ 9x^2 - 126x - 4y^2 + 48y = 279 \]

### Step 2: Factor out the coefficients of the squared terms

\[ 9(x^2 - 14x) - 4(y^2 - 12y) = 279 \]

### Step 3: Complete the square

For \(x\):

\[ x^2 - 14x \]
To complete the square, add and subtract \((\frac{14}{2})^2 = 49\):
\[ x^2 - 14x + 49 - 49 = (x - 7)^2 - 49 \]

For \(y\):

\[ y^2 - 12y \]
To complete the square, add and subtract \((\frac{12}{2})^2 = 36\):
\[ y^2 - 12y + 36 - 36 = (y - 6)^2 - 36 \]

### Step 4: Substitute back into the equation

\[ 9((x - 7)^2 - 49) - 4((y - 6)^2 - 36) = 279 \]
\[ 9(x - 7)^2 - 441 - 4(y - 6)^2 + 144 = 279 \]

### Step 5: Simplify

\[ 9(x - 7)^2 - 4(y - 6)^2 - 297 = 279 \]
\[ 9(x - 7)^2 - 4(y - 6)^2 = 576 \]

### Step 6: Normalize the equation by dividing both sides by 576

\[ \frac{9(x - 7)^2}{576} - \frac{4(y - 6)^2}{576} = 1 \]
\[ \frac{(x - 7)^2}{64} - \frac{(y - 6)^2}{144} = 1 \]

This is the standard form of the equation for a hyperbola:

\[ \frac{(x - 7)^2}{a^2} - \frac{(y - 6)^2}{b^2} = 1 \]

where \(a^2 = 64\) and \(b^2 = 144\).

Thus, the given equation represents a **hyperbola**.

Would you like more details or have any questions about this solution?

Here are 5 related questions you might find helpful:
1. How do you find the asymptotes of a hyperbola?
2. What is the standard form of an ellipse?
3. How can you identify a parabola from its equation?
4. What are the vertices of a hyperbola?
5. How do you find the eccentricity of a hyperbola?

**Tip:** Completing the square is a powerful method to rewrite conic section equations in their standard form, making it easier to identify the type of conic section.

Learn how to identify a conic section using algebraic techniques. This detailed step-by-step solution demonstrates completing the square to classify the equation \( 9x^2 - 4y^2 - 126x + 48y - 279 = 0 \) as a hyperbola.

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