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 \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (where \(a>b>0\)), so it is an ellipse centered at the origin \((0,0)\). Here, \(a^{2}=9\) so \(a = 3\), and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis, since \(a\) is under \(y^{2}\)): The vertices are at \((0,\pm a)=(0,\pm3)\).
• For the \(x\) - axis (minor 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 \((-2,0)\).
• Then draw a smooth ellipse connecting these points. The ellipse will be taller along the \(y\) - axis because the major axis is along the \(y\) - axis (since \(a = 3\) and \(b = 2\), and \(a>b\)).

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

1. Recognize it is an ellipse centered at the origin with \(a = 3\) (along \(y\) - axis) and \(b = 2\) (along \(x\) - axis).
2. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth curve through these points, forming an ellipse that is taller vertically.

(Note: Since the question asks to graph the equation, the key steps are identifying the type of conic, finding the key points, and then plotting them. If we were to describe the graph, it is an ellipse centered at \((0,0)\) with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\), and it is symmetric about both the \(x\) - axis and \(y\) - axis.)

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 \(a>b>0\)), so it is an ellipse centered at the origin \((0,0)\). Here, \(a^{2}=9\) so \(a = 3\), and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis, since \(a\) is under \(y^{2}\)): The vertices are at \((0,\pm a)=(0,\pm3)\).
• For the \(x\) - axis (minor 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 \((-2,0)\).
• Then draw a smooth ellipse connecting these points. The ellipse will be taller along the \(y\) - axis because the major axis is along the \(y\) - axis (since \(a = 3\) and \(b = 2\), and \(a>b\)).

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

1. Recognize it is an ellipse centered at the origin with \(a = 3\) (along \(y\) - axis) and \(b = 2\) (along \(x\) - axis).
2. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth curve through these points, forming an ellipse that is taller vertically.

(Note: Since the question asks to graph the equation, the key steps are identifying the type of conic, finding the key points, and then plotting them. If we were to describe the graph, it is an ellipse centered at \((0,0)\) with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\), and it is symmetric about both the \(x\) - axis and \(y\) - axis.)