Question

graph each equation.

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

graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines, origin at (0,0)

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 \(a>b>0\) and the major axis is along the \(y\)-axis).

Step2: Find the vertices and co - vertices

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\), we have \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

• The vertices (end - points of the major axis) are at \((0,\pm a)=(0,\pm3)\).
• The co - vertices (end - points of the minor axis) are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the points \((0, 3)\), \((0,- 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane.
• Then draw a smooth curve connecting these points to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

To graph the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):

1. Recognize it is an ellipse with major axis along the \(y\) - axis, \(a = 3\), \(b = 2\).
2. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth, symmetric curve through these points. The graph is an ellipse centered at the origin \((0,0)\), 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 \(x = 2\)).