The given equation is:

\[
\frac{(x-6)^2}{36} + \frac{(y-4)^2}{16} = 1
\]

This is the equation of an ellipse in its standard form:

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

Where:
- \((h, k)\) is the center of the ellipse,
- \(a\) is the length of the semi-major axis,
- \(b\) is the length of the semi-minor axis.

From the equation:

- The center \((h, k)\) is \((6, 4)\),
- The value of \(a^2 = 36\), so \(a = 6\),
- The value of \(b^2 = 16\), so \(b = 4\).

### Key features of the ellipse:
1. **Center**: \((6, 4)\),
2. **Semi-major axis**: 6 (along the x-axis, since 36 > 16),
3. **Semi-minor axis**: 4 (along the y-axis),
4. **Orientation**: Since the larger denominator is under the \(x\)-term, the major axis is horizontal,
5. **Vertices**: Located at \((6 \pm 6, 4)\) = \((0, 4)\) and \((12, 4)\),
6. **Co-vertices**: Located at \((6, 4 \pm 4)\) = \((6, 0)\) and \((6, 8)\).

Would you like further details on how to derive any of these values?

Here are 5 related questions:
1. How do you calculate the foci of the ellipse?
2. What is the eccentricity of this ellipse?
3. How does changing the coefficients in the equation affect the shape of the ellipse?
4. How would the equation look if the ellipse were rotated?
5. Can you write the parametric equations for this ellipse?

**Tip:** The larger the difference between \(a\) and \(b\), the more elongated the ellipse will be along its major axis.

Explore how to solve and analyze the equation of an ellipse (x-6)^2/36 + (y-4)^2/16 = 1. Learn about the center, axes, and key properties of ellipses in this problem, suitable for students in Grades 10-12.

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