Question

graph each equation. 9) \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\) graph with x from -8 to 8, y from -8 to 8, grid lines

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

• The vertices of a vertical ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) are at \((0,\pm a)\). So, the vertices are \((0, 3)\) and \((0,- 3)\).
• The co - vertices are at \((\pm b,0)\). So, the co - vertices are \((2,0)\) and \((- 2,0)\).

Step4: Plot the points and draw the ellipse

Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) 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:

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

1. Recognize it is a vertical ellipse centered at \((0,0)\) with \(a = 3\), \(b = 2\).
2. Plot vertices \((0,3)\), \((0,-3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth ellipse through these points. The graph is 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).