SOLUTION: I have a hyperbola that has a center at (0,-9) and one vertex is (-2,-9) and the other lies at (2,-9). Part of the equation reads x(squared)

SOLUTION: I have a hyperbola that has a center at (0,-9) and one vertex is (-2,-9) and the other lies at (2,-9). Part of the equation reads x(squared) - (y+9)=1...but i dont know how to find

Question 32370: I have a hyperbola that has a center at (0,-9) and one vertex is (-2,-9) and the other lies at (2,-9). Part of the equation reads x(squared) - (y+9)=1...but i dont know how to find the dividen or botom half of the equation...i also dont know how to find the answer?
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING EAMPLES AND TRY.
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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
graph( 600, 600, -10, 10, -10, 10, 4*(1+x^2/25)^0.5,-4*(1+x^2/25)^0.5,4*x/5,-4*x/5 )
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What is the vertices, foci, and slope of the asymptote for the hyperbola whose equation is, y^2 - 4x^2 - 2y - 16x + 1 = 0?
(Y^2-2*Y*1+1^2)-{(2X)^2+2*(2X)*4+4^2}-1^2+4^2+1=0
(Y-1)^2-(2X+4)^2=-16
4(X+2)^2-(Y-1)^2=16
{4(X+2)^2}/16-{(Y-1)^2}/16=1
(X+2)^2/2^2-(Y-1)^2/4^2=1...
COMPARING WITH STANDARD EQN.
(X-H)^2/A^2-(Y-K)^2/B^2=1....WE HAVE
VERTICES ARE {(H-A),K} AND {(H+A),K}=(-2-2,1) AND (-2+2,1)=(-4,1) AND (0,1)
FOCI ARE {(H-AE),K} AND {(H+AE),K}...WHERE E IS
ECCENTRICITY =SQRT{(A^2+B^2)/A^2}=SQRT((4+16)/4)=SQRT(5)
SO FOCI ARE =(-2-2SQRT(5),1) AND (-2+2SQRT(5),1)
SLOPE OF ASYMPTOTE IS GIVEN BY DIFFERENTIATION.HAVE YOU BEEN TAUGHT?PLEASE INFORM.I SHALL COME BACK ON HEARING FROM YOU.
or you can take this proposition as proved formula
the pair of asymptotes for a conic is given by the same equation as the conic except for the constant term which has to be found using the condition for the equation to represent a pair of straight lines.
HENCE EQN OF ASYMPTOTES IS GIVEN BY
y^2 - 4x^2 - 2y - 16x + K=0 , WHERE K IS DETERMINED using the condition for the equation to represent a pair of straight lines.
SINCE WE ARE TO FIND ONLY SLOPES ,WE NEED NOT DETERMINE THE CONSTANT BUT ASSUME THAT THIS EQN REPRESENTS A PAIR OF STRAIGHT LINES.SO
y^2 - 4x^2 - 2y - 16x + K=0 = (Y+2X+A)(Y-2X+B)
HENCE SLOPES ARE +2 AND -2
graph( 600, 600, -10, 10, -10, 10, 1-2*((x+2)^2-4)^0.5,1+2*((x+2)^2-4)^0.5,2x+5,-2x-3 )

Quadratic-relations-and-conic-sections/28185: What is the vertices, foci, and slope of the asymptotes for the hyperbola whose equation is, x^2/81 - y^2/36 = 1?
1 solutions
Answer 15969 by venugopalramana(1088) About Me on 2006-03-04 03:09:19 (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
graph( 600, 600, -10, 10, -10, 10, 4*(1+x^2/25)^0.5,-4*(1+x^2/25)^0.5,4*x/5,-4*x/5 )
------------------------------------------
What is the vertices, foci, and slope of the asymptote for the hyperbola whose equation is, y^2 - 4x^2 - 2y - 16x + 1 = 0?
(Y^2-2*Y*1+1^2)-{(2X)^2+2*(2X)*4+4^2}-1^2+4^2+1=0
(Y-1)^2-(2X+4)^2=-16
4(X+2)^2-(Y-1)^2=16
{4(X+2)^2}/16-{(Y-1)^2}/16=1
(X+2)^2/2^2-(Y-1)^2/4^2=1...
COMPARING WITH STANDARD EQN.
(X-H)^2/A^2-(Y-K)^2/B^2=1....WE HAVE
VERTICES ARE {(H-A),K} AND {(H+A),K}=(-2-2,1) AND (-2+2,1)=(-4,1) AND (0,1)
FOCI ARE {(H-AE),K} AND {(H+AE),K}...WHERE E IS
ECCENTRICITY =SQRT{(A^2+B^2)/A^2}=SQRT((4+16)/4)=SQRT(5)
SO FOCI ARE =(-2-2SQRT(5),1) AND (-2+2SQRT(5),1)
SLOPE OF ASYMPTOTE IS GIVEN BY DIFFERENTIATION.HAVE YOU BEEN TAUGHT?PLEASE INFORM.I SHALL COME BACK ON HEARING FROM YOU.
or you can take this proposition as proved formula
the pair of asymptotes for a conic is given by the same equation as the conic except for the constant term which has to be found using the condition for the equation to represent a pair of straight lines.
HENCE EQN OF ASYMPTOTES IS GIVEN BY
y^2 - 4x^2 - 2y - 16x + K=0 , WHERE K IS DETERMINED using the condition for the equation to represent a pair of straight lines.
SINCE WE ARE TO FIND ONLY SLOPES ,WE NEED NOT DETERMINE THE CONSTANT BUT ASSUME THAT THIS EQN REPRESENTS A PAIR OF STRAIGHT LINES.SO
y^2 - 4x^2 - 2y - 16x + K=0 = (Y+2X+A)(Y-2X+B)
HENCE SLOPES ARE +2 AND -2
graph( 600, 600, -10, 10, -10, 10, 1-2*((x+2)^2-4)^0.5,1+2*((x+2)^2-4)^0.5,2x+5,-2x-3 )
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