Get the asymptotes of (y-1)²/9 + (x+3)²/4 = 1

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.

Learn how to identify asymptotes in conic sections. This example explores the equation (y-1)²/9 + (x+3)²/4 = 1, which represents an ellipse. Discover the distinction between ellipses and hyperbolas, and understand why ellipses do not have asymptotes, but hyperbolas do. Suitable for Grades 10-12.

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