Graph (x^2)/64+(y^2)/36=1

Problem

(x^2)/64+(y^2)/36=1

Solution

1. Identify the type of conic section. Since the equation is in the form (x^2)/(a^2)+(y^2)/(b^2)=1 with positive coefficients, it represents an ellipse centered at the origin (0,0)

2. Determine the lengths of the semi-axes. The denominators are a^2=64 and b^2=36 which means a=8 and b=6

3. Locate the vertices and co-vertices. Since a>b the major axis is horizontal. The vertices are at (±8,0) and the co-vertices are at (0,±6)

4. Calculate the focal length c using the relationship c^2=a^2−b^2

c^2=64−36

c^2=28

c=√(,28)=2√(,7)

5. Identify the foci. The foci are located on the major axis at (±2√(,7),0) which is approximately (±5.29,0)

6. Sketch the graph by plotting the center (0,0) the vertices (8,0) and (−8,0) and the co-vertices (0,6) and (0,−6) then drawing a smooth curve through these points.

Final Answer

(x^2)/64+(y^2)/36=1* is an ellipse with vertices *(±8,0)* and co-vertices *(0,±6)

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