Question

graph each equation.

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

Explanation:

🆕 New Concept Discovered: Graphing Ellipses
An oval shape centered at the origin.

Step 1: Identify the conic section

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 \( (0,0) \):

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

Since the denominators are different and both terms are positive, the graph is an ellipse (a stretched circle).

Step 2: Find the intercepts

To find where the ellipse crosses the axes, we look at the denominators:

• Horizontal stretch (\(x\)-direction):

The denominator under \(x^2\) is \(4\), so \(a^2 = 4\), which means \(a = 2\).
This tells us the ellipse goes \(2\) units left and right from the center.
The \(x\)-intercepts are at \( (2, 0) \) and \( (-2, 0) \).

• Vertical stretch (\(y\)-direction):

The denominator under \(y^2\) is \(9\), so \(b^2 = 9\), which means \(b = 3\).
This tells us the ellipse goes \(3\) units up and down from the center.
The \(y\)-intercepts are at \( (0, 3) \) and \( (0, -3) \).

Step 3: Plot the points and draw the curve

1. Plot the center at the origin \( (0,0) \).
2. Plot the vertices along the \(y\)-axis (the major axis because \(9 > 4\)) at \( (0, 3) \) and \( (0, -3) \).
3. Plot the co-vertices along the \(x\)-axis at \( (2, 0) \) and \( (-2, 0) \).
4. Connect these four points with a smooth, symmetric oval curve.

Answer:

To graph the equation \(\frac{x^2}{4} + \frac{y^2}{9} = 1\), plot an ellipse centered at \( (0,0) \) with the following key boundary points:

• Vertices (top and bottom): \( (0, 3) \) and \( (0, -3) \)
• Co-vertices (left and right): \( (2, 0) \) and \( (-2, 0) \)

Draw a smooth oval passing through these four points.