Question

graph each equation.

1. \\(dfrac{x^2}{4} + dfrac{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 \(a>b\) for a vertical major axis). Here, \(a^2 = 9\) so \(a = 3\), and \(b^2=4\) so \(b = 2\). The center of the ellipse is at \((0,0)\) (the origin) since there are no shifts in \(x\) or \(y\) in the equation.

Step2: Find the vertices and co - vertices

• For the major axis (along the \(y\) - axis, since \(a\) is associated with \(y^2\)): The vertices are at \((0,\pm a)=(0,\pm3)\). So the points are \((0, 3)\) and \((0,- 3)\).
• For the minor axis (along the \(x\) - axis): The co - vertices are at \((\pm b,0)=(\pm2,0)\). So the points are \((2,0)\) and \((-2,0)\).

Step3: Plot the points and sketch the ellipse

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\), and the co - vertices \((2,0)\), \((-2,0)\). Then, sketch a smooth ellipse passing through these points. The ellipse will be taller along the \(y\) - axis (since \(a = 3\) is larger than \(b=2\)) and symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

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

1. Recognize it as an ellipse with center \((0,0)\), vertical major axis (since \(a = 3\), \(b = 2\), \(a>b\)), vertices at \((0,\pm3)\), and co - vertices at \((\pm2,0)\).
2. Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\).
3. Draw a smooth, symmetric ellipse through these points. The ellipse is centered at the origin, extends 3 units up and down along the \(y\) - axis, and 2 units left and right along the \(x\) - axis.