Let's analyze and solve the given problem. We are tasked with rewriting the ellipse equation in standard form and then extracting key details.

Step 1: Rewrite the ellipse equation in standard form

The given equation is:

$16x^2 + 9y^2 - 64x - 108y + 244 = 0$

Group and complete the square for $x$ and $y$:

1. Group $x$ terms and $y$ terms:

   $16(x^2 - 4x) + 9(y^2 - 12y) = -244$

2. Complete the square for $x$:

   For $x^2 - 4x$, the coefficient of $x$ is $-4$. Half of $-4$ is $-2$, and squaring it gives $4$.

   Add and subtract $4$ inside the parentheses:

   $16(x^2 - 4x + 4 - 4) = 16((x - 2)^2 - 4)$

   This simplifies to:

   $16(x - 2)^2 - 64$

3. Complete the square for $y$:

   For $y^2 - 12y$, the coefficient of $y$ is $-12$. Half of $-12$ is $-6$, and squaring it gives $36$.

   Add and subtract $36$ inside the parentheses:

   $9(y^2 - 12y + 36 - 36) = 9((y - 6)^2 - 36)$

   This simplifies to:

   $9(y - 6)^2 - 324$

4. Rewrite the equation with completed squares:

   Substituting back, we get:

   $16(x - 2)^2 - 64 + 9(y - 6)^2 - 324 = -244$

   Combine constants:

   $16(x - 2)^2 + 9(y - 6)^2 - 388 = -244$

   Add $388$ to both sides:

   $16(x - 2)^2 + 9(y - 6)^2 = 144$

5. Divide through by 144 to normalize:

   $\frac{16(x - 2)^2}{144} + \frac{9(y - 6)^2}{144} = 1$

   Simplify the denominators:

   $\frac{(x - 2)^2}{9} + \frac{(y - 6)^2}{16} = 1$

Step 2: Identify the ellipse parameters

From the standard form:

$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$

we can identify:

• $h = 2$
• $k = 6$
• $a^2 = 16 \implies a = 4$
• $b^2 = 9 \implies b = 3$

Since $a > b$, the major axis is vertical.

Step 3: Find the endpoints of the axes

Major axis (vertical):

The major axis is along $y = k = 6$. The endpoints are:

$(x, y) = (2, 6 \pm a) = (2, 6 + 4) \text{ and } (2, 6 - 4)$

Endpoints:

$(2, 10) \quad \text{and} \quad (2, 2)$

Minor axis (horizontal):

The minor axis is along $x = h = 2$. The endpoints are:

$(x, y) = (2 \pm b, 6) = (2 + 3, 6) \text{ and } (2 - 3, 6)$

Endpoints:

$(5, 6) \quad \text{and} \quad (-1, 6)$

Step 4: Foci of the ellipse

The foci lie along the major axis at a distance of $c$ from the center, where:

$c = \sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7}$

Thus, the foci are:

$(x, y) = (2, 6 \pm \sqrt{7})$

Final Answer:

• Standard form: $\frac{(x - 2)^2}{9} + \frac{(y - 6)^2}{16} = 1$
• Center ($h, k$): $(2, 6)$
• $a = 4$, $b = 3$
• Major axis endpoints: $(2, 10)$ and $(2, 2)$
• Minor axis endpoints: $(5, 6)$ and $(-1, 6)$
• Foci: $(2, 6 + \sqrt{7})$ and $(2, 6 - \sqrt{7})$

Would you like further clarification on any step?

Explore Further:

1. What is the formula for the eccentricity of an ellipse?
2. How do we determine whether an ellipse is horizontal or vertical?
3. Can the minor axis ever be longer than the major axis?
4. How does completing the square help in deriving the standard form of conic sections?
5. What is the geometric meaning of the foci?

Tip: Always confirm whether $a^2$ or $b^2$ is larger to correctly identify the major axis of an ellipse!