Question

graph each equation. 9) \\(\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 \(a>b>0\) 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 \(y\) - axis (major axis): 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 vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((- 2,0)\) on the coordinate plane.
• Then, sketch the ellipse by connecting these points smoothly, making sure that the ellipse 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 as an ellipse with \(a = 3\) (along \(y\) - axis) and \(b = 2\) (along \(x\) - axis).
2. Plot vertices \((0,3)\), \((0,-3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Sketch the ellipse through these points, symmetric about \(x\) and \(y\) axes.