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Question 48302This question is from textbook College Algebra
: I need help, I can't seem to grasp this concept. The question is.. identify the center, vertices, foci, and equations of the asymptotes of the hyperbola: 25y^2 - 9x^2 - 100y - 54x - 206 = 0. If you can help it would be greatly appreciated. Thank you lots.
This question is from textbook College Algebra
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I need help, I can't seem to grasp this concept. The question is.. identify the center, vertices, foci, and equations of the asymptotes of the hyperbola: 25y^2 - 9x^2 - 100y - 54x - 206 = 0. If you can help it would be greatly appreciated.
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FIRST COMPLETE SQUARES
25(Y^2-4y)-9(X^2+6X)-206=0
25{Y^2-2*Y*2+2^2-2^2}-9{X^2+2*X*3+3^2-3^2}-206=0
25(Y-2)^2-100-9(X+3)^2+81-206=0
-9(X+3)^2+25(Y-2)^2=-81+100+206=225
-[{(X+3)^2}/(225/9)]+[{(Y-2)^2}/(225/25)]=1
[(Y-2)^2/5^2]-[(X+3)^2/3^2]=1
NOW
SEE THE FOLLOWING AND YOU SHOULD BE ABLE TO SOLVE YOUR PROBLEM BY YOUR SELF..
Quadratic-relations-and-conic-sections/28184: What is the vertices,
foci, and slope of the asymptotes for the hyperbola whose equation is,
y^2/16 - x^2/25 =25?
1 solutions
Answer 15970 by venugopalramana(1088) About Me on 2006-03-04 03:12:07
(Show Source):
SEE THE FOLLOWING AND YOU SHOULD BE ABLE TO SOLVE YOUR PROBLEM BY YOUR SELF.....
THE ANSWERS FOR YOUR CASE...H=0...K=0..A=4...B=5....EQN IS OF THE TYPE
(Y-K)^2/B^2-(X-H)^2/A^2=1....
SO VERTICES ARE...(H,(K-B)) AND (H,(K+B)) ...(0,-5) AND (0,5)
FOCI ARE {H,(K-BE)} AND {H,(K+BE)}...WHERE E IS
ECCENTRICITY =SQRT{(A^2+B^2)/B^2}=SQRT((16+25)/25)=SQRT(41/25)
SO FOCI ARE =(0,-5SQRT(41/25) AND (0,5SQRT(41/25)...
OR....(0,-SQRT(41)) AND (0,SQRT(41)
ASYMPTOTES ARE GIVEN BY
Y^2/16-X^2/25=K
25Y^2-16X^2-400K=0
(5Y+4X+A)(5Y-4X+B)=0
SLOPES OF ASYMPTOTES ARE
-4/5 AND 4/5
THE GRAPHS LOOK LIKE THIS
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