Part 1: Completing the square to rewrite the equation in standard form:

We start with the given equation: $49x^2 + 64y^2 - 294x + 128y - 2631 = 0$

Step 1: Group $x$- and $y$-terms:

$49(x^2 - 6x) + 64(y^2 + 2y) = 2631$

Step 2: Complete the square for $x$-terms:

$x^2 - 6x \to (x - 3)^2 - 9$ $49((x - 3)^2 - 9) = 49(x - 3)^2 - 441$

Step 3: Complete the square for $y$-terms:

$y^2 + 2y \to (y + 1)^2 - 1$ $64((y + 1)^2 - 1) = 64(y + 1)^2 - 64$

Step 4: Substitute back and simplify:

$49(x - 3)^2 - 441 + 64(y + 1)^2 - 64 = 2631$ $49(x - 3)^2 + 64(y + 1)^2 = 2631 + 441 + 64$ $49(x - 3)^2 + 64(y + 1)^2 = 3136$

Step 5: Divide through by 3136:

$\frac{(x - 3)^2}{64} + \frac{(y + 1)^2}{49} = 1$

Thus, the standard form of the equation is: $\boxed{\frac{(x - 3)^2}{64} + \frac{(y + 1)^2}{49} = 1}$

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Part 2: Graph and foci:

This is an ellipse with a horizontal major axis because the larger denominator is under the $(x - 3)^2$-term. Key details:

• Center: $(3, -1)$
• Semi-major axis: $\sqrt{64} = 8$
• Semi-minor axis: $\sqrt{49} = 7$

Foci:

The distance from the center to each focus is: $c = \sqrt{a^2 - b^2} = \sqrt{64 - 49} = \sqrt{15} \approx 3.87$ The foci are located horizontally from the center: $(3 \pm \sqrt{15}, -1) \approx (3 \pm 3.87, -1)$ $\text{Foci: } (6.87, -1) \text{ and } (-0.87, -1)$

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Part 3: Correct graph:

The ellipse has a horizontal major axis, is centered at $(3, -1)$, and passes through points like $(3 + 8, -1) = (11, -1)$ and $(3 - 8, -1) = (-5, -1)$, making option C correct.

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Summary:

1. Standard form: $\frac{(x - 3)^2}{64} + \frac{(y + 1)^2}{49} = 1$
2. Foci: $(6.87, -1)$ and $(-0.87, -1)$
3. Correct graph: C

Let me know if you’d like further clarifications!

Related Questions:

1. How do you determine the foci of an ellipse?
2. What is the difference between a horizontal and vertical major axis?
3. Can you explain how to calculate eccentricity for an ellipse?
4. How do you derive the length of axes from an ellipse equation?
5. What role does completing the square play in standardizing equations?

Tip: Always check which denominator is larger to quickly identify the major axis!