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Question 28101: 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)?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 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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