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Question 28461: What is the center, foci, and the length of the major and minor axes for the ellipse whose equation is, (x-4)^2/16 + (y+1)^2/9 = 1?
Answer by venugopalramana(3286)   (Show Source):
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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
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