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Question 28102: What is the center, foci, and the lengths of the major and minor axes for the ellipse, x^2/5 + y^2/10 = 1?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLES AND FOLLOW THE SAME METHOD TO SOLVE YOUR PROBLEM.IF STILL IN DIFFICULTY COME BACK.
What is the equation for the ellipse, the foci are at (12,0) and (-12,0). The endpoints of the minor axis are at (0,5)and (0,-5)?
1 solutions
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Answer 15328 by venugopalramana(965) on 2006-02-22 00:32:29 (Show Source):
What is the equation for the ellipse, the foci are at (12,0) and (-12,0). The endpoints of the minor axis are at (0,5)and (0,-5)?
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STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 + (Y-K)^2/B^2 = 1 ,FOR A>B, WHOSE FOCI ARE (H+A*E,K)AND (H-A*E,K)WHERE
E=SQRT{(A^2-B^2)/A^2} AND MINOR AXIS IS ALONG X=H
AND LENGTH OF MAJOR AXIS =2A AND LENGTH OF MINOR AXIS =2B
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WE FIND X COORDINATES OF END POINTS OF MINOR AXIS ARE 0.HENCE MINOR AXIS IS Y AXIS.
FURTHER WE FIND Y COORDINATES OF FOCI ARE 0...HENCE MAJOR AXIS IS X AXIS
HENCE A>B
HENCE CENTRE IS (0,0).SO H=K=0
LENGTH OF MINOR AXIS = 5+5 = 10 = 2B...OR B=5
SO NOW WE NEED TO FIND ONLY A TO GET THE EQN.OF ELLIPSE.
WE GOT FOCI AS (12,0)=(0+A*E,0)
AE=12....OR E=12/A
BUT
E=SQRT{(A^2-B^2)/A^2} =12/A...SQUARING...
144/A^2=(A^2-B^2)/A^2
A^2-B^2=144
A^2=144+5^2=144+25=169
A=13
HENCE EQN. OF ELLIPSE IS
X^2/169 + Y^2/25 = 1
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Quadratic-relations-and-conic-sections/28099: Would you please help me name the center of the ellipse x^2/64 + (y-3)^2/36 = 1 and then help me state whether the major axis is horizontal or vertical?
1 solutions
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Answer 15327 by venugopalramana(965) on 2006-02-22 00:12:30 (Show Source):
Would you please help me name the center of the ellipse x^2/64 + (y-3)^2/36 = 1 and then help me state whether the major axis is horizontal or vertical?
COMPARE WITH STANDARD EQN.
(X-H)^2/A^2 + (Y-K)^2 = 1 WHOSE CENTRE IS (H,K),WE GET CENTRE AS (0,3)
SINCE A=8 IS GREATER THAN B=6,MAJOR AXIS IS ALONG X AXIS OR HORIZONTAL.
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