Question

graph each equation. 9) $\frac{x^2}{4} + \frac{y^2}{9} = 1$

Explanation:

Step1: Identify the conic section

The equation \(\frac{x^2}{4}+\frac{y^2}{9} = 1\) is in the standard form of an ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (since \(a^2 = 9\) and \(b^2=4\), and \(a>b\), so it is a vertical ellipse). For an ellipse of the form \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\), the center is at \((0,0)\), the vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).

Step2: Calculate the values of \(a\) and \(b\)

Given \(a^2=9\), then \(a = \sqrt{9}=3\). Given \(b^2 = 4\), then \(b=\sqrt{4} = 2\).

Step3: Find the key points

• The center of the ellipse is \((h,k)=(0,0)\) (since there are no shifts in the \(x\) and \(y\) terms).
• The vertices (the endpoints of the major axis) are at \((0, a)=(0,3)\) and \((0, - a)=(0, - 3)\).
• The co - vertices (the endpoints of the minor axis) are at \((b,0)=(2,0)\) and \((-b,0)=(- 2,0)\).

Step4: Plot the points and draw the ellipse

1. Plot the center \((0,0)\).
2. Plot the vertices \((0,3)\) and \((0, - 3)\).
3. Plot the co - vertices \((2,0)\) and \((-2,0)\).
4. Then, draw an ellipse passing through these four points, making sure it is symmetric about both the \(x\) - axis and \(y\) - axis.

(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with center at the origin, vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above. If we were to describe the graph in words, it is an ellipse centered at \((0,0)\) that is taller along the \(y\) - axis, with the top and bottom points at \((0,3)\) and \((0, - 3)\) and the left and right points at \((-2,0)\) and \((2,0)\).)

Answer:

The graph is an ellipse centered at \((0,0)\) with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\) (plotted and connected on the given coordinate grid).