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

Problem

(x2)/25+(y2)/16=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 represents an ellipse centered at the origin (0,0)

2. Determine the lengths of the semi-axes. Here, a2=25 and b2=16 which means a=5 and b=4 Since a>b the major axis is horizontal along the xaxis.

3. Locate the vertices and co-vertices. The vertices are at (±a,0) which are (5,0) and (−5,0) The co-vertices are at (0,±b) which are (0,4) and (0,−4)

4. Calculate the foci using the relationship c2=a2−b2 Substituting the values gives c2=25−16=9 so c=3 The foci are located at (±3,0)

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

Final Answer

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

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