Question

graph each equation.

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

Explanation:

Step1: Identify the conic section type

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which 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 \((0,0)\). The vertices (end - points of the major axis) are at \((0,\pm a)\) and the co - vertices (end - points of the minor axis) are at \((\pm b,0)\).
Since \(a^{2}=9\), then \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\).
Since \(b^{2}=4\), then \(b = 2\), so the co - vertices are \((2,0)\) and \((-2,0)\).

Step3: Plot the center, vertices, and co - vertices

• The center of the ellipse is at the origin \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\) (these are the points on the \(y\) - axis, 3 units above and below the origin).
• Plot the co - vertices \((2,0)\) and \((-2,0)\) (these are the points on the \(x\) - axis, 2 units to the right and left of the origin).

Step4: Sketch the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

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

1. Locate the center at \((0,0)\).
2. Mark the vertices at \((0, 3)\) and \((0,-3)\) (along the \(y\) - axis, 3 units from the center).
3. Mark the co - vertices at \((2,0)\) and \((-2,0)\) (along the \(x\) - axis, 2 units from the center).
4. Draw a smooth curve connecting these points, making sure the ellipse is symmetric about both the \(x\) - axis and \(y\) - axis.

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

Answer:

Step1: Identify the conic section type

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which 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 \((0,0)\). The vertices (end - points of the major axis) are at \((0,\pm a)\) and the co - vertices (end - points of the minor axis) are at \((\pm b,0)\).
Since \(a^{2}=9\), then \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\).
Since \(b^{2}=4\), then \(b = 2\), so the co - vertices are \((2,0)\) and \((-2,0)\).

Step3: Plot the center, vertices, and co - vertices

• The center of the ellipse is at the origin \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\) (these are the points on the \(y\) - axis, 3 units above and below the origin).
• Plot the co - vertices \((2,0)\) and \((-2,0)\) (these are the points on the \(x\) - axis, 2 units to the right and left of the origin).

Step4: Sketch the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

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

1. Locate the center at \((0,0)\).
2. Mark the vertices at \((0, 3)\) and \((0,-3)\) (along the \(y\) - axis, 3 units from the center).
3. Mark the co - vertices at \((2,0)\) and \((-2,0)\) (along the \(x\) - axis, 2 units from the center).
4. Draw a smooth curve connecting these points, making sure the ellipse is symmetric about both the \(x\) - axis and \(y\) - axis.

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