Question

graph each equation.

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

graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines, origin at (0,0)

Explanation:

Step1: Identify the ellipse standard form

The given equation is \(\frac{x^2}{4}+\frac{y^2}{9} = 1\), which matches the standard form of an ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (where \(a > b\) and the major axis is vertical). Here, \(a^2=9\) so \(a = 3\), and \(b^2 = 4\) so \(b=2\).

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 \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).
• Vertices: Substitute \(a = 3\) into \((0,\pm a)\), we get \((0,3)\) and \((0, - 3)\).
• Co - vertices: Substitute \(b = 2\) into \((\pm b,0)\), we get \((2,0)\) and \((- 2,0)\).

Step3: Plot the points and sketch the ellipse

Plot the points \((0,3)\), \((0,-3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane. Then, sketch a smooth ellipse passing through these points. The ellipse will be centered at the origin \((0,0)\) (since there are no \(h\) or \(k\) shifts in the standard form \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\), here \(h = 0\) and \(k = 0\)), with a vertical major axis (because \(a^2\) is under the \(y^2\) term) of length \(2a=6\) and a horizontal minor axis of length \(2b = 4\).

Answer:

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

1. Identify it as a vertical ellipse centered at \((0,0)\) with \(a = 3\) (vertical radius) and \(b=2\) (horizontal radius).
2. Plot vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Sketch 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.