Question

graph each equation.

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

Explanation:

Step1: Identify the conic type

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 \(a > b\), it is a vertical - major - axis ellipse). Here, \(a=\sqrt{9}=3\) and \(b = \sqrt{4}=2\).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\):

• The vertices are at \((0,\pm a)=(0,\pm3)\).
• The co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the vertices \((0, 3)\), \((0,- 3)\) and the co - vertices \((2,0)\), \((-2,0)\).
• Then, sketch the ellipse by connecting these points smoothly, making sure that the ellipse is symmetric about both the \(x\) - axis and \(y\) - axis.

To graph the ellipse:

1. Mark the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane.
2. Draw a smooth curve passing through these points, symmetric about the \(x\) - axis and \(y\) - axis. The major axis is along the \(y\) - axis with length \(2a = 6\) and the minor axis is along the \(x\) - axis with length \(2b=4\).

Answer:

The graph is an ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\), plotted and connected smoothly on the given coordinate grid. (The actual graph is a smooth curve passing through \((0, 3)\), \((0,-3)\), \((2,0)\), \((-2,0)\) and symmetric about the axes.)