Find the Properties ((x-2)^2)/49+((y-2)^2)/40=1

Problem

((x−2)2)/49+((y−2)2)/40=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 with a2>b2 it is a horizontal ellipse.

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

h=2

k=2

Center=(2,2)

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

a2=49⇒a=7

b2=40⇒b=√(,40)=2√(,10)

4. Find the distance to the foci c using the relationship c2=a2−b2

c2=49−40=9

c=3

5. Locate the vertices by adding and subtracting a from the xcoordinate of the center.

(V_1)=(2+7,2)=(9,2)

(V_2)=(2−7,2)=(−5,2)

6. Locate the foci by adding and subtracting c from the xcoordinate of the center.

(F_1)=(2+3,2)=(5,2)

(F_2)=(2−3,2)=(−1,2)

7. Locate the co-vertices by adding and subtracting b from the ycoordinate of the center.

C*(V_1)=(2,2+2√(,10))

C*(V_2)=(2,2−2√(,10))

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

e=3/7

Final Answer

Center: *(2,2), Vertices: *(9,2),(−5,2), Foci: *(5,2),(−1,2), Eccentricity: 3/7

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