(x-1)^2/4 + (y-1)^2/12 = 1 show step by step process solving of the equation of asymptotes in hyperbola 

The given equation is:

\[
\frac{(x - 1)^2}{4} + \frac{(y - 1)^2}{12} = 1
\]

This equation is not a hyperbola but rather an ellipse. Let's first explain the steps in identifying the nature of the conic section and how to find the asymptotes if it were a hyperbola. Then, we will proceed with the correction for this particular case.

### Step-by-Step Explanation:

#### Step 1: General Form of Conic Sections
- The standard form of an **ellipse** is:
  \[
  \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
  \]
  - If \(a^2 > b^2\), the ellipse is wider horizontally.
  - If \(b^2 > a^2\), the ellipse is taller vertically.

- The standard form of a **hyperbola** is:
  \[
  \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \quad \text{(horizontal hyperbola)}
  \]
  or
  \[
  \frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2} = 1 \quad \text{(vertical hyperbola)}
  \]

In your case, since both terms are positive, the equation describes an **ellipse**.

#### Step 2: Identify Parameters from the Equation
Your given equation is:
\[
\frac{(x - 1)^2}{4} + \frac{(y - 1)^2}{12} = 1
\]
Comparing this to the standard form of an ellipse:
- \( (h, k) = (1, 1) \) is the center of the ellipse.
- \( a^2 = 4 \), so \( a = 2 \), which represents the semi-major axis (horizontal).
- \( b^2 = 12 \), so \( b = 2\sqrt{3} \), which represents the semi-minor axis (vertical).

#### Step 3: Focus and Asymptotes (if it were a hyperbola)
For an ellipse, there are **no asymptotes** because an ellipse is bounded (unlike a hyperbola). However, if the equation had been for a hyperbola, the asymptotes would follow from:

For a hyperbola of the form:
\[
\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
\]
The equations of the asymptotes are given by:
\[
y - k = \pm \frac{b}{a}(x - h)
\]

For the equation \(\frac{(x - 1)^2}{4} - \frac{(y - 1)^2}{12} = 1\) (hyperbola):
- The slopes of the asymptotes would be \(\pm \frac{b}{a} = \pm \frac{\sqrt{12}}{2} = \pm \sqrt{3}\).
- The asymptote equations would then be:
  \[
  y - 1 = \pm \sqrt{3}(x - 1)
  \]
  
### Conclusion
The given equation is an **ellipse** and does not have asymptotes. However, if it had been a hyperbola, we would follow the method above to calculate the asymptotes. 

Do you need any additional details or further clarifications on this problem?

### Follow-up Questions:
1. What is the standard form of a hyperbola, and how does it differ from an ellipse?
2. How do we determine whether a conic is a hyperbola or an ellipse from its equation?
3. What role do the parameters \(a\) and \(b\) play in determining the shape of ellipses and hyperbolas?
4. How can we find the foci of a hyperbola or an ellipse?
5. Can an ellipse have asymptotes under any conditions?

### Tip:
Always check the signs between terms when identifying the type of conic section!

(x-1)^2/4 + (y-1)^2/12 = 1 show step by step process solving of the equation of asymptotes in hyperbola

Understand how to differentiate between an ellipse and a hyperbola in the equation (x-1)^2/4 + (y-1)^2/12 = 1. Explore the key formulas for conic sections, step-by-step solutions, and why the given equation represents an ellipse without asymptotes. Suitable for Grades 10-12.

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