Find the Properties ((x+2)^2)/64+((y-1)^2)/81=1

Problem

((x+2)^2)/64+((y−1)^2)/81=1

Solution

1. Identify the type of conic section. Since the equation is in the form ((x−h)^2)/(b^2)+((y−k)^2)/(a^2)=1 with a^2>b^2 this is a vertical ellipse.

2. Determine the center (h,k) by looking at the terms (x+2) and (y−1)

h=−2

k=1

Center=(−2,1)

3. Calculate the semi-major axis a and semi-minor axis b

a^2=81⇒a=9

b^2=64⇒b=8

4. Find the distance to the foci c using the relationship c^2=a^2−b^2

c^2=81−64=17

c=√(,17)

5. Locate the vertices by shifting the center vertically by a

Vertices=(−2,1±9)

Vertex 1=(−2,10)

Vertex 2=(−2,−8)

6. Locate the foci by shifting the center vertically by c

Foci=(−2,1±√(,17))

7. Locate the co-vertices by shifting the center horizontally by b

Co-vertices=(−2±8,1)

Co-vertex 1=(6,1)

Co-vertex 2=(−10,1)

8. Calculate the eccentricity e using the formula e=c/a

e=√(,17)/9

Final Answer

Center: *(−2,1), Vertices: *(−2,10),(−2,−8), Foci: *(−2,1±√(,17)), Eccentricity: √(,17)/9

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