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 \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (since \(a^2 = 9\) and \(b^2=4\), and \(a>b\), so it is a vertical ellipse).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\), the center is at \((0,0)\) (since there are no shifts in \(x\) or \(y\) from the origin).

• The length of the semi - major axis \(a=\sqrt{9} = 3\), so the vertices are at \((0, a)=(0,3)\) and \((0, - a)=(0, - 3)\).
• The length of the semi - minor axis \(b=\sqrt{4}=2\), so the co - vertices are at \((b,0)=(2,0)\) and \((-b,0)=(- 2,0)\).

Step3: Plot the points and draw the ellipse

Plot the center \((0,0)\), vertices \((0,3)\), \((0, - 3)\), and co - vertices \((2,0)\), \((-2,0)\). Then, sketch the ellipse by connecting these points smoothly, making sure that the ellipse is 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 a vertical ellipse with center \((0,0)\).
2. Plot vertices \((0, 3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth ellipse passing through these points, symmetric about the \(x\) - axis and \(y\) - axis.