SOLUTION: Given the ellipse: 2x^2 + 6y^2 + 32x - 48y + 212 = 0 find. a. The Center C b. Length of Major Axis c. Length of Minor Axis d. Distance from C to foci

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Given the ellipse: 2x^2 + 6y^2 + 32x - 48y + 212 = 0 find. a. The Center C b. Length of Major Axis c. Length of Minor Axis d. Distance from C to foci      Log On


   

Question 259707: Given the ellipse: 2x^2 + 6y^2 + 32x - 48y + 212 = 0 find.
a. The Center C
b. Length of Major Axis
c. Length of Minor Axis
d. Distance from C to foci
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
We have this original equation as
(i) 2x%5E2+%2B+6y%5E2+%2B+32x+-+48y+%2B+212+=+0
rearranging, we get
(ii) %282x%5E2+%2B+32x%29+%2B+%286y%5E2+-+48y%29+=+-212
complete the square to get
(iii) 2%28x%2B8%29%5E2+%2B+6%28y-4%29%5E2+=+-212+%2B+128+%2B96
to get
(iv) 2%28x%2B8%29%5E2+%2B+6%28y-4%29%5E2+=+12
dividing by 12, we get
(v) %28%28x%2B8%29%5E2%29%2F6+%2B+%28%28y-4%29%5E2%29%2F2+=+1
---
center is (-8,4)
--
major axis = sqrt(6)
--
minor axis = sqrt(2)
--
distance to foci
sqrt(sqrt(6) - sqrt(2)) = sqrt(4) = 2