Find the Properties ((x+6)^2)/12+((y-4)^2)/16=1

Problem

((x+6)^2)/12+((y−4)^2)/16=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+6) and (y−4)

h=−6

k=4

Center=(−6,4)

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

a^2=16⇒a=4

b^2=12⇒b=√(,12)=2√(,3)

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

c^2=16−12=4

c=2

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

Vertices=(−6,4±4)

(Vertex_1)=(−6,8)

(Vertex_2)=(−6,0)

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

Foci=(−6,4±2)

(Focus_1)=(−6,6)

(Focus_2)=(−6,2)

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

Co-vertices=(−6±2√(,3),4)

8. Determine the lengths of the axes.

Major Axis Length=2*a=8

Minor Axis Length=2*b=4√(,3)

9. Calculate the eccentricity e

e=c/a=2/4=0.5

Final Answer

Center: *(−6,4), Vertices: *(−6,8),(−6,0), Foci: *(−6,6),(−6,2), Eccentricity: *0.5

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