Question

graph each equation.

1. \\(\dfrac{x^2}{4} + \dfrac{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 \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (since \(a^{2}=9\) and \(b^{2} = 4\), and \(a>b\), 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 (end - points of the major axis) are at \((0,\pm a)\). Since \(a=\sqrt{9}=3\), the vertices are \((0, 3)\) and \((0,-3)\).
• The co - vertices (end - points of the minor axis) are at \((\pm b,0)\). Since \(b = \sqrt{4}=2\), 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, sketch a smooth ellipse passing through these points. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

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

1. Recognize it as a vertical ellipse centered at \((0,0)\) with \(a = 3\) (semi - major axis) and \(b=2\) (semi - minor axis).
2. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth ellipse through these points, symmetric about the \(x\) and \(y\) axes. The graph will be an ellipse that is taller along the \(y\) - axis, with its top at \((0,3)\), bottom at \((0, - 3)\), rightmost point at \((2,0)\) and leftmost point at \((-2,0)\).