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

Problem

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

Solution

1. Identify the type of conic section. The equation is in the standard form of an ellipse centered at the origin (0,0) which is (x^2)/(a^2)+(y^2)/(b^2)=1 or (x^2)/(b^2)+(y^2)/(a^2)=1

2. Determine the values of a and b Since 9>4 the ellipse is vertical. We have a^2=9 which means a=3 and b^2=4 which means b=2

3. Find the vertices and co-vertices. The vertices are located at (0,±a) which are (0,3) and (0,−3) The co-vertices are located at (±b,0) which are (2,0) and (−2,0)

4. Calculate the foci using the formula c^2=a^2−b^2

c^2=9−4

c^2=5

c=√(,5)

The foci are located at (0,±√(,5))

5. Sketch the graph by plotting the center (0,0) the vertices, and the co-vertices, then drawing a smooth curve through the points to form the ellipse.

Final Answer

(x^2)/4+(y^2)/9=1* is an ellipse with vertices *(0,±3)* and co-vertices *(±2,0)

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