Graph ((x-2)^2)/16+((y+1)^2)/9=1

Problem

((x−2)2)/16+((y+1)2)/9=1

Solution

1. Identify the type of conic section. Since the equation is in the form ((x−h)2)/(a2)+((y−k)2)/(b2)=1 it represents an ellipse.

2. Determine the center (h,k) of the ellipse. By comparing the given equation to the standard form, we find h=2 and k=−1

C*e*n*t*e*r=(2,−1)

3. Calculate the lengths of the semi-major and semi-minor axes. Here, a2=16 and b2=9 so a=4 and b=3 Since a>b the major axis is horizontal.

4. Locate the vertices by moving a units left and right from the center.

(2±4,−1)⇒(6,−1)* and *(−2,−1)

5. Locate the co-vertices by moving b units up and down from the center.

(2,−1±3)⇒(2,2)* and *(2,−4)

6. Find the foci using the relationship c2=a2−b2

c2=16−9=7

c=√(,7)≈2.65

F*o*c*i=(2±√(,7),−1)

7. Sketch the graph by plotting the center, vertices, and co-vertices, then drawing a smooth oval through the four outer points.

Final Answer

((x−2)2)/16+((y+1)2)/9=1* is an ellipse centered at *(2,−1)* with vertices at *(6,−1),(−2,−1)* and co-vertices at *(2,2),(2,−4)

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