Question

graph each equation.

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

coordinate plane with x from -8 to 8 and y from -8 to 8, grid lines, axes labeled x and y

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 centered at the origin \((0,0)\)).

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 \((0,\pm a)\). Since \(a^{2}=9\), then \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\).
• The co - vertices are at \((\pm b,0)\). Since \(b^{2}=4\), then \(b = 2\), so the co - vertices are \((2,0)\) and \((-2,0)\).

Step3: Plot the points and draw the ellipse

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\) on the coordinate plane. Then, draw a smooth curve connecting these points to form the ellipse. 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 centered at the origin with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\), and the ellipse is drawn by connecting these points smoothly. (To actually draw it, plot the points and sketch the ellipse passing through them.)