Graph (x^2)/16+(y^2)/25=1

Problem

(x2)/16+(y2)/25=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 semi-axes lengths. Comparing the equation to the standard form, we find a2=16 and b2=25 which gives a=4 and b=5

3. Locate the vertices and co-vertices. Since b>a the major axis is vertical. The vertices are at (0,±5) and the co-vertices are at (±4,0)

4. Calculate the foci using the relationship c2=|a2−b2|

c2=25−16

c2=9

c=3

5. Identify the foci coordinates. The foci are located along the major axis (y-axis) at (0,±3)

6. Sketch the graph by plotting the center (0,0) the vertices (0,5) and (0,−5) and the co-vertices (4,0) and (−4,0) then drawing a smooth curve through these points.

Final Answer

(x2)/16+(y2)/25=1* is an ellipse with vertices *(0,±5)* and co-vertices *(±4,0)

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