SOLUTION: 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.

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 Click here to see ALL problems on Quadratic-relations-and-conic-sections 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)