Question

graph each equation.

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

coordinate grid with x from -8 to 8 and y from -8 to 8, axes labeled x and y

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 \((0,0)\) (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 \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\). Then, sketch the ellipse passing through these points, making sure it is symmetric about both the \(x\) - axis and \(y\) - axis.

To graph the ellipse:

1. Mark the center at the origin \((0,0)\).
2. Move 3 units up and down from the center to mark the vertices \((0,3)\) and \((0, - 3)\).
3. Move 2 units left and right from the center to mark the co - vertices \((-2,0)\) and \((2,0)\).
4. Draw a smooth curve connecting these points, forming an ellipse that is taller along the \(y\) - axis.

(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with center at \((0,0)\), vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above.)

Answer:

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 \((0,0)\) (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 \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\). Then, sketch the ellipse passing through these points, making sure it is symmetric about both the \(x\) - axis and \(y\) - axis.

To graph the ellipse:

1. Mark the center at the origin \((0,0)\).
2. Move 3 units up and down from the center to mark the vertices \((0,3)\) and \((0, - 3)\).
3. Move 2 units left and right from the center to mark the co - vertices \((-2,0)\) and \((2,0)\).
4. Draw a smooth curve connecting these points, forming an ellipse that is taller along the \(y\) - axis.

(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with center at \((0,0)\), vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above.)