Find the Properties ((x+4)^2)/121+((y-8)^2)/196=1

Problem

((x+4)^2)/121+((y−8)^2)/196=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 it is a vertical ellipse.

2. Determine the center (h,k) by looking at the terms (x+4) and (y−8)

h=−4

k=8

Center=(−4,8)

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

a^2=196⇒a=14

b^2=121⇒b=11

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

c^2=196−121=75

c=√(,75)=5√(,3)

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

(V_1)=(−4,8+14)=(−4,22)

(V_2)=(−4,8−14)=(−4,−6)

6. Locate the co-vertices by adding and subtracting b from the xcoordinate of the center.

C*(V_1)=(−4+11,8)=(7,8)

C*(V_2)=(−4−11,8)=(−15,8)

7. Locate the foci by adding and subtracting c from the ycoordinate of the center.

(F_1)=(−4,8+5√(,3))

(F_2)=(−4,8−5√(,3))

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

e=(5√(,3))/14

Final Answer

Center: *(−4,8), Vertices: *(−4,22),(−4,−6), Foci: *(−4,8±5√(,3)), Eccentricity: (5√(,3))/14

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