graph each equation.\n9) $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

graph each equation.\n9) $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$
Answer
Explanation:
Step1: Identify the conic section type
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) (since (a^{2}=9) and (b^{2}=4), and (a > b)), so it is an ellipse centered at the origin ((0,0)) with a vertical major axis.
Step2: Find the vertices and co - vertices
For the ellipse (\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1) ((a>b>0)):
- The length of the semi - major axis (a=\sqrt{9} = 3), so the vertices are at ((0,\pm a)=(0, 3)) and ((0,-3)).
- The length of the semi - minor axis (b=\sqrt{4}=2), so the co - vertices are at ((\pm b,0)=(\pm2,0)).
Step3: Plot the points
Plot the center ((0,0)), 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 (\frac{x^{2}}{4}+\frac{y^{2}}{9}=1):
- Recognize it is an ellipse centered at ((0,0)) with vertical major axis.
- Plot vertices ((0,3)), ((0, - 3)) and co - vertices ((2,0)), ((-2,0)).
- Draw a smooth ellipse through these points.
(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with center at the origin, vertices at ((0,\pm3)) and co - vertices at ((\pm2,0)) as described above.)