Question

graph each equation.

1. $dfrac{x^{2}}{4} + dfrac{y^{2}}{9} = 1$

Explanation:

Step1: Identify the conic section

The equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which 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 are at \((0,\pm a)\). Since \(a=\sqrt{9} = 3\), the vertices are \((0, 3)\) and \((0,- 3)\).
• The co - vertices are at \((\pm b,0)\). Since \(b=\sqrt{4}=2\), the co - vertices are \((2,0)\) and \((- 2,0)\).

Step3: Plot the points

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\). Then sketch the ellipse passing through these points.

Answer:

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

1. Recognize it is a vertical ellipse centered at \((0,0)\) with \(a = 3\) (semi - major axis along y - axis) and \(b=2\) (semi - minor axis along x - axis).
2. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw an ellipse through these points. The graph will be an ellipse symmetric about both the x - axis and y - axis, with the top and bottom points at \((0,3)\) and \((0, - 3)\), and the left and right points at \((-2,0)\) and \((2,0)\).