https://file.aiquickdraw.com/mathbot/file/1725892340028i5zcqcus.jpg

Let's break down and solve the problem based on the equation given in the image:

The given equation is:
\[
144x^2 + 81y^2 + 2016x - 648y = 3312
\]
This equation describes an ellipse, and we need to solve the questions related to it.

### Step 1: Rewrite the equation of the ellipse
We start by grouping the \(x\) and \(y\) terms to complete the square:

\[
144x^2 + 2016x + 81y^2 - 648y = 3312
\]

Now, factor out the coefficients of \(x^2\) and \(y^2\) from the grouped terms:
\[
144(x^2 + 14x) + 81(y^2 - 8y) = 3312
\]

### Step 2: Complete the square for \(x\) and \(y\)
For \(x^2 + 14x\), take half of 14 (which is 7) and square it (49), then add and subtract 49 inside the parentheses:
\[
144(x^2 + 14x + 49 - 49)
= 144((x + 7)^2 - 49)
\]
Similarly, for \(y^2 - 8y\), take half of 8 (which is 4) and square it (16), then add and subtract 16 inside the parentheses:
\[
81(y^2 - 8y + 16 - 16)
= 81((y - 4)^2 - 16)
\]

### Step 3: Substitute the completed squares
Now substitute these back into the equation:
\[
144((x + 7)^2 - 49) + 81((y - 4)^2 - 16) = 3312
\]
Simplify:
\[
144(x + 7)^2 - 144 \cdot 49 + 81(y - 4)^2 - 81 \cdot 16 = 3312
\]
\[
144(x + 7)^2 + 81(y - 4)^2 - 7056 - 1296 = 3312
\]
\[
144(x + 7)^2 + 81(y - 4)^2 = 3312 + 7056 + 1296
\]
\[
144(x + 7)^2 + 81(y - 4)^2 = 11664
\]

### Step 4: Divide by 11664 to put in standard form
\[
\frac{144(x + 7)^2}{11664} + \frac{81(y - 4)^2}{11664} = 1
\]
Simplify the fractions:
\[
\frac{(x + 7)^2}{81} + \frac{(y - 4)^2}{144} = 1
\]

This is the standard form of the ellipse equation:
\[
\frac{(x + 7)^2}{9^2} + \frac{(y - 4)^2}{12^2} = 1
\]

### Answer to Question 5:
The standard form is:
\[
\frac{(x + 7)^2}{9^2} + \frac{(y - 4)^2}{12^2} = 1
\]
So, the correct option is **a**.

### Answer to Question 6:
The center of the ellipse is at \((-7, 4)\), since the equation is in the form \((x + h)^2/a^2 + (y - k)^2/b^2 = 1\), where the center is \((h, k)\).

So, the correct option is **d**.

### Answer to Question 7:
The major axis is along the \(y\)-axis because the larger denominator (144) is under the \(y\)-term. The length of the major axis is \(2b = 2 \times 12 = 24\), so the endpoints are 12 units above and below the center. Thus, the endpoints are at \((-7, 4 + 12) = (-7, 16)\) and \((-7, 4 - 12) = (-7, -8)\).

So, the correct option is **a**.

### Answer to Question 8:
The distance between the center and the foci is given by \(c = \sqrt{b^2 - a^2} = \sqrt{144 - 81} = \sqrt{63} \approx 7.94\). The foci are along the major axis, which is vertical, so the coordinates of the foci are at \((-7, 4 + 7.94)\) and \((-7, 4 - 7.94)\), approximately \((-7, 11.94)\) and \((-7, -3.94)\).

So, the correct option is **b**.

---

Do you want any further details or explanations for these solutions? 

Here are 5 related questions to consider:
1. How do you find the eccentricity of this ellipse?
2. How does the length of the major axis affect the shape of the ellipse?
3. What would happen if \( a \) and \( b \) were equal in an ellipse equation?
4. How can you derive the equation of a hyperbola from an ellipse?
5. Can you graph the equation of this ellipse by hand?

**Tip:** To easily spot the center of a conic section like an ellipse, always look for the terms in the form \((x + h)^2\) and \((y - k)^2\) in the equation.

What is the standard form of the equation? What are the coordinates of the center, the endpoints of the major axis, and the coordinates of the two foci?

This page details the solution to finding the standard form, center, major axis, and foci of an ellipse. Includes step-by-step guidance on completing the square and calculating key points. Ideal for Grade 11-12 math students.

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