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Question 32650: Find the equation of a hyperbola with vertices at (0,±5) and asymtotes y=±2x.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! ASYMPTOTES ARE
Y=2X
OR
.Y-2X=0
I
..AND
Y=-2X
OR
.Y+2X=0
II
...SO THEIR COMBINED EQN.IS GIVEN BY EQN.I*EQN.II=0
(Y-2X)(Y+2X)=0=Y^2-4X^2=0
EQN.3.
IT IS KNOWN THAT THE EQN.OF HYPERBOLA DIFFERS FROM THE COMBINED EQN.OF ASYMPTOTES ONLY BY A CONSTANT.
SO
EQN.OF HYPERBOLA IS
.
Y^2-4X^2=C
..III
...WHERE C IS A ONSTANT TO BE FOUND
PUTTING THIS EQN.IN STD.FORM WE GET
Y^2/C -4X^2/C =1
.OR
..
(Y-0)^2/C - (x-0)^2/(C/4) =1
IV
COMPARING WITH STANDARD EQN.OF HYPERBOLA
(Y-K)^2/B^2-(X-H)^2/A^2=1.
V
WHERE VERTICES ARE...(H,(K-B)) AND (H,(K+B))
GIVEN HERE AS (0,-5) AND (0,5)
WE GET H=0
K+B=5
K-B=-5
ADDING THE ABOVE 2 EQNS
..2K=0
OR
K=0
SO
B=5
COMPARING EQN.IV,WITH STD.EQN.V,WE GET
.
H=0
..K=0
.C=B^2
..C/4=A^2
.BUT B=5 WE FOUND .SO
C=5^2=25
.A^2=25/4
HENCE EQN.OF HYPERBOLA IS AS PER EQN.III IS
Y^2-4X^2=25
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