Question

graph each equation.

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

Explanation:

Step1: Identify the conic section

The 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 \(a>b\) for vertical major axis), where \(a^{2} = 9\) and \(b^{2}=4\). So \(a = 3\) and \(b = 2\).

Step2: Find the vertices and co - vertices

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices (endpoints of the major axis) are at \((0,\pm a)=(0,\pm3)\) and the co - vertices (endpoints of the minor axis) are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the vertices \((0, 3)\), \((0,- 3)\) and the co - vertices \((2,0)\), \((-2,0)\).
• Then draw a smooth curve connecting these points to form the ellipse.

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

1. Recognize it is an ellipse with vertical major axis (since the denominator under \(y^{2}\) is larger).
2. Determine \(a = 3\) (from \(a^{2}=9\)) and \(b = 2\) (from \(b^{2}=4\)).
3. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
4. Draw a smooth ellipse passing through these points.

The graph will be an ellipse centered at the origin \((0,0)\), with the top and bottom points at \((0,3)\) and \((0, - 3)\) and the left and right points at \((-2,0)\) and \((2,0)\).

Answer:

The graph is an ellipse centered at \((0,0)\) with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\) (the actual drawing involves plotting these points and connecting them with a smooth curve).