SOLUTION: Write the equation of the specified ellipses: 1) Vertices = (7,0) and (-7,0) and Foci = (6,0) and (-6,0) 2) Foci = (3,0) and (-3,0) and major arc is length of 12 Thank you.

Question 36675: Write the equation of the specified ellipses:
1) Vertices = (7,0) and (-7,0) and Foci = (6,0) and (-6,0)
2) Foci = (3,0) and (-3,0) and major arc is length of 12
Thank you.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Write the equation of the specified ellipses:
1) Vertices = (7,0) and (-7,0) and Foci = (6,0) and (-6,0)
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 + (Y-K)^2/B^2 = 1
WHERE
CENTRE IS (H,K)............WE HAVE CENTRE AS {(7-7)/2,(0-0)/2}=(0,0)..HENCE H=K=0
ECCENTRICITY = E = {(A^2-B^2)/A^2}^0.5 IF A>B....AND (B^2-A^2)/A^2...IF B>A
FOCI ARE (H+AE,K)AND (H-AE,K)...IF A>B ..AND...(H,K+BE) & (H,K-BE)..IF B>A
......WE FIND HERE FOCI HAVE SAME Y COORDINATES..HENCE 0+AE=6....0-AE=-6..OR AE=6
MAJOR AXIS LENGTH = 2A.......OR .....2B WHICHEVER IS BIGGER
HERE MAJOR AXIS =DISTANCE BETWEEN VERTICES =SQRT{(7+7)^2+0)}=14=2A...A=7
MINOR AXIS LENGTH = 2B.......OR......2A WHICHEVER IS SMALLER
AE=6...A=7...HENCE E=6/7
HENCE E=6/7={(A^2-B^2)/A^2}^0.5...SQUARING
36/49=(49-B^2)/49
36=49-B^2
B^2=49-36=13
HENCE EQN. OF ELLIPSE IS
X^2/49+Y^/13=1
2) Foci = (3,0) and (-3,0) and major arc is length of 12
Thank you.....IT SHOULD BE MAZOR AXIS IS OF LENGTH 12.
PROCEEDING SAME WAY AS ABOVE
CENTRE ={3-3)/2,(0-0)/2}=(0,0)...H=K=0
2A=12...A=6
0+AE=3....0-AE=-3...AE=3
E=3/6=1/2
E^2=1/4=(A^2-B^2)/A^2=(36-B^2)/36
36-B^2=36/4=9
B^2=36-9=27
HENCE EQN.OF ELLIPSE IS
X^2/36 +Y^2/27=1
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