Question

graph each equation.

1. \\(\frac{x^2}{4} + \frac{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). Here, \(a=\sqrt{9}=3\) and \(b=\sqrt{4} = 2\). The center of the ellipse is at the origin \((0,0)\) because there are no shifts in the \(x\) or \(y\) terms.

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\):

• The vertices are at \((0,\pm a)\). Substituting \(a = 3\), the vertices are \((0,3)\) and \((0, - 3)\).
• The co - vertices are at \((\pm b,0)\). Substituting \(b=2\), the co - vertices are \((2,0)\) and \((- 2,0)\).

Step3: Plot the points and draw the ellipse

1. Plot the center \((0,0)\).
2. Plot the vertices \((0,3)\) and \((0,-3)\) (these are the points on the \(y\) - axis, 3 units above and below the center).
3. 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 center).
4. Then, draw a smooth curve connecting these points to form the ellipse. The ellipse will be taller along the \(y\) - axis (since \(a>b\)) and wider along the \(x\) - axis with the given radii.

To graph the ellipse:

• Mark the center at \((0,0)\).
• Mark the vertices \((0,3)\) and \((0, - 3)\).
• Mark the co - vertices \((2,0)\) and \((-2,0)\).
• Draw a smooth ellipse passing through these four points.

(Note: Since the question is about graphing, 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). Here, \(a=\sqrt{9}=3\) and \(b=\sqrt{4} = 2\). The center of the ellipse is at the origin \((0,0)\) because there are no shifts in the \(x\) or \(y\) terms.

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\):

• The vertices are at \((0,\pm a)\). Substituting \(a = 3\), the vertices are \((0,3)\) and \((0, - 3)\).
• The co - vertices are at \((\pm b,0)\). Substituting \(b=2\), the co - vertices are \((2,0)\) and \((- 2,0)\).

Step3: Plot the points and draw the ellipse

1. Plot the center \((0,0)\).
2. Plot the vertices \((0,3)\) and \((0,-3)\) (these are the points on the \(y\) - axis, 3 units above and below the center).
3. 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 center).
4. Then, draw a smooth curve connecting these points to form the ellipse. The ellipse will be taller along the \(y\) - axis (since \(a>b\)) and wider along the \(x\) - axis with the given radii.

To graph the ellipse:

• Mark the center at \((0,0)\).
• Mark the vertices \((0,3)\) and \((0, - 3)\).
• Mark the co - vertices \((2,0)\) and \((-2,0)\).
• Draw a smooth ellipse passing through these four points.

(Note: Since the question is about graphing, 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.)