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.
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) (Show Source): You can put this solution on YOUR website! --------------------------------------------------------
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
(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)