Question 46053This question is from textbook College Algebra
: I need help finding the center, vertices, and foci for the following equation:
(x+3)^2/9 + (y+1)^2/16 = 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 need help finding the center, vertices, and foci for the following equation:
(x+3)^2/9 + (y+1)^2/16 = 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=-3.....K=-1 HENCE CENTRE OF ELLIPSE IS AT
IS (-3,-1)
A=SQRT(9)=3........B=SQRT(16)=4
WHERE MAJOR AXIS =2B=2*4=8...MINOR AXIS =2A=2*3=6
ECCENTRICITY =E =SQRT{(B^2-A^2)/B^2}=SQRT{(16-9)/16}=SQRT(7)/4
FOCI ARE GIVEN BY....(H,K+BE),(H,K-BE)
(-3,-1+4*SQRT(7)/4),(-3,-1-4*SQRT(7)/4)={-3,-1+SQRT(7)},{-3,-1-SQRT(7)}
VERTICES ARE (H,K+B),(H,K-B)=(-3,3),(-3,-5)
AND....(H+A,K),(H-A,K)=(0,-1),(-6,-1)
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