How do you find the coordinates of the center, foci, the length of the major and minor axis given (x+8)^2/144+(y-2)^2/81=1(x+8)2144+(y2)281=1?

1 Answer
May 18, 2017

See the answer below

Explanation:

(x+8)^2/(12^2)+(y-2)^2/(9^2)=1(x+8)2122+(y2)292=1

We compare this equation to the standard equation of a horizontal ellipse

(x-h)^2/a^2+(y-k)^2/b^2=1(xh)2a2+(yk)2b2=1

The center of the ellipse is (h,k)=(-8,2)(h,k)=(8,2)

The length of the major axis is =2a=2*12=24=2a=212=24

The length of the major axis is =2b=2*9=18=2b=29=18

We need cc to determine the foci

c^2=a^2-b^2=144-81=53c2=a2b2=14481=53

c=+-sqrt53c=±53

The foci are F=(c,2)=(-8+sqrt53,2)F=(c,2)=(8+53,2) and

F'=(-8-sqrt53,2)
graph{((x+8)^2/144+(y-2)^2/81-1)((x+8)^2+(y-2)^2-0.1)=0 [-26.1, 5.95, -5.07, 10.95]}