Question

graph each equation.

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

coordinate grid with x from -8 to 8 and y from -8 to 8

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).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the center is at \((0,0)\) (the origin).

• The length of the semi - major axis \(a=\sqrt{9}=3\), so the vertices are at \((0,\pm a)=(0, 3)\) and \((0,- 3)\).
• The length of the semi - minor axis \(b = \sqrt{4}=2\), so the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((- 2,0)\). Then draw a smooth curve connecting these points to form the ellipse.

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

1. Recognize it is a vertical ellipse with center \((0,0)\).
2. Mark the vertices \((0,3)\), \((0, - 3)\) (since \(a = 3\)) and co - vertices \((2,0)\), \((-2,0)\) (since \(b=2\)).
3. Draw a smooth elliptical curve passing through these points. The ellipse will be taller along the \(y\) - axis (because the major axis is along the \(y\) - axis) with a width of \(4\) units (from \(x=-2\) to \(x = 2\)) and a height of \(6\) units (from \(y=-3\) to \(y = 3\)) centered at the origin.

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 a smooth curve connecting these points (a vertical ellipse with semi - major axis length \(3\) and semi - minor axis length \(2\)).