Question

graph each equation.

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

graph with x-axis from -8 to 8 and y-axis 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 \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (since \(a > b\) and the major axis is along the \(y\)-axis), where \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• The vertices of the ellipse (endpoints of the major axis) are at \((0,\pm a)=(0,\pm3)\).
• The co - vertices (endpoints of the minor axis) are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the points \((0, 3)\), \((0,- 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane.
• Then, draw an 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 the denominator of \(y^{2}\) is larger) 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. Recognize it as an 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 passing through these four points. The ellipse is taller than it is wide, centered at the origin, with the top and bottom points at \((0,3)\) and \((0, - 3)\) and the left and right points at \((-2,0)\) and \((2,0)\).