Question

graph each equation.

1. \\( \frac { x ^ { 2 } } { 4 } + \frac { 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 \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (where \(a>b>0\)), which is a vertical ellipse centered at the origin \((0,0)\).

Step2: Find the values of \(a\) and \(b\)

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

Step3: Determine the vertices and co - vertices

• Vertices: For a vertical ellipse centered at the origin, the vertices are at \((0,\pm a)\). Substituting \(a = 3\), the vertices are \((0,3)\) and \((0, - 3)\).
• Co - vertices: For a vertical ellipse centered at the origin, the co - vertices are at \((\pm b,0)\). Substituting \(b = 2\), the co - vertices are \((2,0)\) and \((- 2,0)\).

Step4: 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 the ellipse passing through these points. 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:

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

1. Recognize it is a vertical ellipse centered at \((0,0)\) with \(a = 3\) (semi - major axis) and \(b=2\) (semi - minor axis).
2. Plot vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw an ellipse through these points with the major axis along the \(y\) - axis and minor axis along the \(x\) - axis.

(The graph will be an ellipse centered at the origin, stretching 3 units up and down from the center along the \(y\) - axis and 2 units left and right from the center along the \(x\) - axis.)