Question

graph each equation.

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

Explanation:

Step1: Identify the ellipse standard form

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 \(a > b\) for vertical major axis), where \(a^2 = 9\) 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 \(y\). Substitute \(x = 0\) into the equation \(\frac{0^2}{4}+\frac{y^2}{9}=1\), then \(y^2=9\), so \(y=\pm3\). The vertices are \((0, 3)\) and \((0,- 3)\).
• For the \(x\) - axis (minor axis) co - vertices: When \(y = 0\), we solve for \(x\). Substitute \(y = 0\) into the equation \(\frac{x^2}{4}+\frac{0^2}{9}=1\), then \(\frac{x^2}{4}=1\), \(x^2 = 4\), so \(x=\pm2\). The co - vertices are \((2, 0)\) and \((- 2,0)\).

Step3: Plot the points

• Plot the vertices \((0,3)\), \((0, - 3)\) and 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 question is to graph the equation, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above. If we were to describe the key points for graphing: vertices \((0, 3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\) are plotted and the ellipse is drawn through them.)

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\) (and the ellipse is drawn passing through these points).