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
