Graph (x^2)/9+(y^2)/4=1

Problem

(x2)/9+(y2)/4=1

Solution

1. Identify the type of conic section. Since the equation is in the form (x2)/(a2)+(y2)/(b2)=1 with positive coefficients, it is an ellipse centered at the origin (0,0)

2. Determine the lengths of the semi-axes. Here, a2=9 and b2=4 which means a=3 and b=2

3. Locate the vertices on the x-axis. Since a=3 is under the x2 term, the horizontal vertices are at (3,0) and (−3,0)

4. Locate the vertices on the y-axis. Since b=2 is under the y2 term, the vertical vertices (co-vertices) are at (0,2) and (0,−2)

5. Calculate the foci using the formula c2=a2−b2 This gives c2=9−4=5 so c=√(,5) The foci are located at (√(,5),0) and (−√(,5),0)

6. Sketch the graph by drawing a smooth curve through the four vertices (3,0) (−3,0) (0,2) and (0,−2)

Final Answer

(x2)/9+(y2)/4=1* is an ellipse with vertices at *(±3,0)* and *(0,±2)

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