SOLUTION: I am trying to find the center, vertices, and foci of the ellipse of the following equation: x^2/49 + y^2/36 = 1 Any help will be greatly appreciated.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I am trying to find the center, vertices, and foci of the ellipse of the following equation: x^2/49 + y^2/36 = 1 Any help will be greatly appreciated.      Log On


   

Question 46052This question is from textbook College Algebra
: I am trying to find the center, vertices, and foci of the ellipse of the following equation:
x^2/49 + y^2/36 = 1
Any help will be greatly appreciated. This question is from textbook College Algebra

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
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I am trying to find the center, vertices, and foci of the ellipse of the following equation:
x^2/49 + y^2/36 = 1
STANDARD EQN.OF ELLIPSE IS
Solved by pluggable solver: Linear System solver (using determinant)
Solve:
+system%28+%0D%0A++++1%5Cx+%2B+1%5Cy+=+2%2C%0D%0A++++1%5Cx+%2B+-1%5Cy+=+0+%29%0D%0A++

Any system of equations:


has solution

or



(x=1, y=1}

(X-H)^2/A^2 +(Y-K)^2/B^2=1
CENTRE IS (H,K)..AS PER THE PROBLEM H=0.....K=0 HENCE CENTRE OF ELLIPSE IS AT
IS (0,0)
A=SQRT(49)=7........B=SQRT(36)=6
WHERE MAJOR AXIS =2A=2*7=14...MINOR AXIS =2B=2*6=12
ECCENTRICITY =E =SQRT{(A^2-B^2)/A^2}=SQRT{(49-36)/49}=SQRT(13)/7
FOCI ARE GIVEN BY....(H+AE,K),(H-AE,K)
(SQRT(13),0),(-SQRT(13),0)
VERTICES ARE (H,K+B),(H,K-B)=(0,6),(0,-6)
AND....(H+A,K),(H-A,K)=(7,0),(-7,0)