The given equation is:

$\frac{(y-1)^2}{9} + \frac{(x+3)^2}{4} = 1$

This is the equation of an ellipse in standard form:

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

where $(h, k)$ is the center of the ellipse, $a^2$ is the square of the semi-major axis, and $b^2$ is the square of the semi-minor axis.

Step 1: Identifying the Center, Semi-Major Axis, and Semi-Minor Axis

The center $(h, k)$ is $(-3, 1)$.

• $a^2 = 9$, so $a = 3$.
• $b^2 = 4$, so $b = 2$.

Since the $y$-term is associated with the larger denominator ($9$), the major axis is vertical.

Step 2: Determining the Asymptotes

Asymptotes only exist for hyperbolas, not ellipses. Since the equation you provided represents an ellipse, there are no asymptotes for this equation.

However, if you intended to ask about a hyperbola with a similar form, the equation of a hyperbola would look like:

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

Would you like further clarification or more information? Here are some related questions:

1. How do you distinguish between ellipses and hyperbolas?
2. What are the steps to derive the asymptotes of a hyperbola?
3. Can an ellipse have slant asymptotes?
4. How would the equation change if it were a hyperbola instead of an ellipse?
5. How do we graph ellipses and identify their axes?

Tip: Asymptotes are characteristic of hyperbolas and define their end-behavior, while ellipses are closed curves with no asymptotes.