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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Would you like a detailed explanation on any part of the solution?

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.