SOLUTION: What is the center, foci, and the lengths of the major and minor axes for the ellipse, 16x^2 + 25y^2 + 32x - 150y = 159?

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Question 28183: What is the center, foci, and the lengths of the major and minor axes for the
ellipse, 16x^2 + 25y^2 + 32x - 150y = 159?
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE
What is the center, foci, and the length of the major and minor axes for the ellipse whose equation is, 16x^2 + 25y^2 + 32x - 150y = 159?
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 + (Y-K)^2/B^2 = 1
WHERE
CENTRE IS (H,K)
ECCENTRICITY = E = {(A^2-B^2)/A^2}^0.5
FOCI ARE (H+AE,K)AND (H-AE,K)
MAJOR AXIS LENGTH = 2A
MINOR AXIS LENGTH = 2B
WE HAVE
16x^2 + 25y^2 + 32x - 150y - 159 =0
{(4X)^2+2*(4X)*4+4^2}-4^2+{(5Y)^2-2*(5Y)*15+15^2}-15^2-159=0
(4X+4)^2 + (5Y-15)^2 = 400
16(X+1)^2 + 25(Y-3)^2 =400...DIVIDING BY 400 THROUGHOUT..
(X+1)^2/25 + (Y-3)^2/16 =1
(X+1)^2/5^2 + (Y-3)^2/4^2 =1
COMPARING WITH ABOVE STANDARD EQN.
CENTRE IS (-1,3)
ECCENTRICITY E IS {(25-16)/25}^0.5=3/5
FOCI ARE (-1+5*3/5 ,3) AND (-1-5*3/5,3)=(2,3) AND (-4,-3)
MAJOR AXIS LENGTH = 2*5=10
MINOR AXIS LENGTH = 2*4=8