Question

graph each equation.

1. \\(\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 \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (where \(a > b\) for a vertical major axis). Here, \(a^2=9\) so \(a = 3\) and \(b^2 = 4\) so \(b=2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis) vertices: When \(x = 0\), we solve for \(y\). Substituting \(x = 0\) into the equation \(\frac{0^2}{4}+\frac{y^2}{9}=1\), we get \(\frac{y^2}{9}=1\), so \(y^2=9\) and \(y=\pm3\). So the vertices are \((0, 3)\) and \((0,- 3)\).
• For the \(x\) - axis (minor axis) co - vertices: When \(y = 0\), we solve for \(x\). Substituting \(y = 0\) into the equation \(\frac{x^2}{4}+\frac{0^2}{9}=1\), we get \(\frac{x^2}{4}=1\), so \(x^2 = 4\) and \(x=\pm2\). So the co - vertices are \((2,0)\) and \((- 2,0)\).

Step3: Plot the points

• Plot the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\) on the coordinate plane.
• Then draw a smooth ellipse passing through these points. The major axis is along the \(y\) - axis with length \(2a=6\) and the minor axis is along the \(x\) - axis with length \(2b = 4\).

(Note: Since the problem is to graph the equation, the final answer is the graph of the ellipse with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\) as described above. If we were to describe the graph in words, it's an ellipse centered at the origin, with a vertical major axis of length 6 (from \(y=-3\) to \(y = 3\)) and a horizontal minor axis of length 4 (from \(x=-2\) to \(x = 2\)).)

Answer:

The graph is an ellipse centered at the origin \((0,0)\), with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\) (the ellipse is drawn through these points with the major axis along the \(y\) - axis and minor axis along the \(x\) - axis).