To solve the equation \( 4x^2 + 9y^2 - 16x - 54y + 61 = 0 \), we will first attempt to rewrite it into the standard form of an ellipse by completing the square for both \(x\) and \(y\).

### Step 1: Group the \(x\)-terms and the \(y\)-terms
\[
4x^2 - 16x + 9y^2 - 54y = -61
\]

### Step 2: Factor out the coefficients of \(x^2\) and \(y^2\)
\[
4(x^2 - 4x) + 9(y^2 - 6y) = -61
\]

### Step 3: Complete the square for \(x\) and \(y\)
- For \(x^2 - 4x\): Take half of \(-4\), square it to get \((-4/2)^2 = 4\).
- For \(y^2 - 6y\): Take half of \(-6\), square it to get \((-6/2)^2 = 9\).

Now add these constants inside the parentheses, and adjust the equation accordingly:
\[
4(x^2 - 4x + 4) + 9(y^2 - 6y + 9) = -61 + 4(4) + 9(9)
\]
\[
4(x - 2)^2 + 9(y - 3)^2 = -61 + 16 + 81
\]
\[
4(x - 2)^2 + 9(y - 3)^2 = 36
\]

### Step 4: Divide through by 36 to get the standard form of an ellipse
\[
\frac{4(x - 2)^2}{36} + \frac{9(y - 3)^2}{36} = \frac{36}{36}
\]
\[
\frac{(x - 2)^2}{9} + \frac{(y - 3)^2}{4} = 1
\]

### Final Equation:
\[
\frac{(x - 2)^2}{9} + \frac{(y - 3)^2}{4} = 1
\]

This is the equation of an ellipse centered at \((2, 3)\), with a semi-major axis of 3 units along the \(x\)-axis and a semi-minor axis of 2 units along the \(y\)-axis.

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Here are 5 related questions:
1. How do you find the foci of an ellipse?
2. Can this equation represent a hyperbola? If so, how would the steps change?
3. What is the general form of a conic section?
4. How do you determine the axes lengths of an ellipse from its equation?
5. What are the applications of ellipses in physics and engineering?

**Tip:** In any conic section, completing the square is a critical step in transforming the equation into its standard form.

Learn how to solve the conic section equation 4x^2 + 9y^2 - 16x - 54y + 61 = 0 by completing the square to express it as an ellipse. Understand key concepts in algebra and conic sections, including ellipses, completing the square, and deriving the standard form. Suitable for Grades 10-12 students.

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