Question

graph each equation.

1. \frac{x^2}{4} + \frac{y^2}{9} = 1

Explanation:

Identify the conic section type

The given equation is:

$$\frac{x^2}{4} + \frac{y^2}{9} = 1$$

This matches the standard form of an ellipse centered at the origin \((0,0)\):

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

Determine the semi-axes lengths

Compare the denominators to find the values of \(a\) and \(b\):

$$a^2 = 4 \implies a = 2$$

$$b^2 = 9 \implies b = 3$$

Since \(b > a\), the ellipse is vertically oriented.

Find the key vertices

The vertices along the horizontal major/minor axis are:

$$(\pm a, 0) = (\pm 2, 0)$$

The vertices along the vertical major/minor axis are:

$$(0, \pm b) = (0, \pm 3)$$

Plot the key points

Mark these four boundary points on the coordinate grid:

• Right vertex: \((2, 0)\)
• Left vertex: \((-2, 0)\)
• Top vertex: \((0, 3)\)
• Bottom vertex: \((0, -3)\)

Sketch the ellipse

Draw a smooth, symmetric oval curve passing through the four plotted points:

• \((2, 0)\)
• \((-2, 0)\)
• \((0, 3)\)
• \((0, -3)\)

Answer:

The equation represents a vertical ellipse centered at the origin \((0,0)\).

• Center: \((0,0)\)
• Horizontal intercepts (vertices): \((2, 0)\) and \((-2, 0)\)
• Vertical intercepts (vertices): \((0, 3)\) and \((0, -3)\)