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!