Question

graph each equation.

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

graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines, origin at (0,0)

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 \(a>b>0\) and the major axis is along the \(y\)-axis). Here, \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the major axis (along \(y\)-axis), the vertices are at \((0,\pm a)=(0,\pm3)\).
• For the minor axis (along \(x\)-axis), the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the points \((0, 3)\), \((0,- 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane.

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be centered at the origin \((0,0)\), with a vertical major axis (since \(a>b\) and the larger denominator is under the \(y^{2}\) term) passing through \((0,\pm3)\) and horizontal minor axis passing through \((\pm2,0)\).

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

1. Identify the type of conic: This is an ellipse with the standard form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\) (vertical major axis, \(a = 3\), \(b=2\), center at \((0,0)\)).
2. Plot key points:

• Vertices (along \(y\)-axis): \((0, 3)\) and \((0,-3)\).
• Co - vertices (along \(x\)-axis): \((2, 0)\) and \((-2, 0)\).

1. Draw the ellipse: Connect these points with a smooth, closed curve.

(Note: Since the question asks to graph the equation, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) centered at the origin. If we were to describe the graph in terms of its key features, the ellipse is centered at \((0,0)\), has a vertical major axis of length \(6\) (from \(y=- 3\) to \(y = 3\)) and a horizontal minor axis of length \(4\) (from \(x=-2\) to \(x = 2\)).)

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 \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (where \(a>b>0\) and the major axis is along the \(y\)-axis). Here, \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the major axis (along \(y\)-axis), the vertices are at \((0,\pm a)=(0,\pm3)\).
• For the minor axis (along \(x\)-axis), the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the points \((0, 3)\), \((0,- 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane.

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be centered at the origin \((0,0)\), with a vertical major axis (since \(a>b\) and the larger denominator is under the \(y^{2}\) term) passing through \((0,\pm3)\) and horizontal minor axis passing through \((\pm2,0)\).

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

1. Identify the type of conic: This is an ellipse with the standard form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\) (vertical major axis, \(a = 3\), \(b=2\), center at \((0,0)\)).
2. Plot key points:

• Vertices (along \(y\)-axis): \((0, 3)\) and \((0,-3)\).
• Co - vertices (along \(x\)-axis): \((2, 0)\) and \((-2, 0)\).

1. Draw the ellipse: Connect these points with a smooth, closed curve.

(Note: Since the question asks to graph the equation, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) centered at the origin. If we were to describe the graph in terms of its key features, the ellipse is centered at \((0,0)\), has a vertical major axis of length \(6\) (from \(y=- 3\) to \(y = 3\)) and a horizontal minor axis of length \(4\) (from \(x=-2\) to \(x = 2\)).)