Question

graph each equation.

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

Explanation:

Step1: Identify the ellipse standard form

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 \(a > b\) and major axis is along y - axis) where \(a^{2}=9\) and \(b^{2}=4\). So, \(a = 3\) and \(b = 2\).

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 vertices \((0, 3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
• Then, sketch the ellipse passing through these four points, making sure it is symmetric about both the x - axis and y - axis.

Answer:

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

1. Recognize it is an ellipse with major axis along the \(y\) - axis, \(a = 3\) (semi - major axis), \(b = 2\) (semi - minor axis).
2. Plot vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw an ellipse through these points, symmetric about \(x\) and \(y\) axes. The graph will be an ellipse centered at the origin \((0,0)\) with the top - most point at \((0,3)\), bottom - most at \((0, - 3)\), right - most at \((2,0)\) and left - most at \((-2,0)\).