SOLUTION: Hyperbolas, i need to write an equations for a hyperbola centered at (-4,5) and has a vertical transverse axis. the value of "a" is 4 and the value of "b" is 9. help me

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Question 27828: Hyperbolas,
i need to write an equations for a hyperbola centered at (-4,5) and has a vertical transverse axis. the value of "a" is 4 and the value of "b" is 9.
help me please!!!
Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website!
SEE FOLLOWING TO UNDERSTAND ASYMPTOTES AND THE AXES

YOU WILL FIND 2 SYMMETRIC CURVES LYING HORIZONTALLY..THEY ARE A PAIR OF HYPERBOLAS .THEIR EQN.IS
(X^2/9)-(Y^2/9)=1...THE GENERAL EQN. IS (X^2/A^2)-(Y^2/B^2)=1
THE OTHER PAIR OF HYPERBOLAS WHICH ARE IN A VERTICAL POSITION ARE CALLED CONJUGATE HYPERBOLAS OF THE EARLIER 2 HYPERBOLAS.THEIR EQN.IS
(X^2/9)-(Y^2/9)= -1...THE GENERAL EQN. IS (X^2/A^2)-(Y^2/B^2)= -1
NOW WE DEFINE THE VARIOUS TERMS WITH RESPECT TO THE FIRST HYPERBOLAS GIVEN BY
(X^2/9)-(Y^2/9)=1...THE GENERAL EQN. IS (X^2/A^2)-(Y^2/B^2)=1
DEFINITIONS.......
HYPERBOLA IS THE LOCUS (OR PATH TRACED)BY A POINT WHICH MOVES SUCH THAT ITS DISTANCE FROM A FIXED POINT CALLED FOCUS TO ITS DISTANCE FROM A FIXED LINE CALLED DIRECTRIX IS CONSTANT (KNOWN AS ECCENTRICITY)AND IS MORE THAN 1.
THERE ARE 2 FOCI AND 2 DIRECTIXES FOR THE HYPERBOLA GIVEN BY THE ABOVE DEFINITION AND EQN.
X AXIS ALONG WHICH THE 2 HYPERBOLAS LIE IS CALLED TRANSVERSE AXIS.ITS EQN.IS Y=0
Y AXIS ALONG WHICH THE 2 CONJUGATE HYPERBOLAS ARE PRESENT IS CALLED THE CONJUGATE AXIS.ITS EQN.IS X=0
IF WE CALL THE 2 POINTS ON EITHER SIDE OF ORIGIN ON THE TRANSVERSE AXIS AT DISTANCE OF A FROM THE ORIGIN ARE NAMED A AND A' THEN AA'=2A IS THE LENGTH OF TRANVERSE AXIS
IF WE CALL THE 2 POINTS ON EITHER SIDE OF ORIGIN ON THE CONJUGATE AXIS AT DISTANCE OF B FROM THE ORIGIN ARE NAMED B AND B' THEN BB'=2B IS THE LENGTH OF CONJUGATE AXIS
ORIGIN IS THE CENTRE OF THE HYPERBOLAS
ECCENTRICITY OF HYPERBOLA IS GIVEN BY E={(A^2+B^2)/(A^2)}^0.5
FOCI ARE GIVEN BY (A/E,0)AND(-A/E,0)
EQNS.OF DIRECTRIX 1 AND 2 ARE GIVEN BY X=A/E AND X=-A/E.
THE 2 LINES YOU FIND DIAGONALLY ALMOST RUNNING PARALLEL TO THE CURVES AT THEIR ENDS ARE CALLED ASYMPTOTES.THE CURVES APPROACH THESE LINES AS NEAR AS WE DESIRE AT AS FAR A DISTANCE AS NEEDED,BUT NEVER TOUCH THEM .THEY RUN PARALLEL AS THE CURVES AND THE LINES EXTEND TO INFINITY .
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GIVEN BELOW ARE SOME MORE EXAMPLES.NOW YOU TRY TO DRAW YOUR REQUIRED CURVE AND UNDERSTAND.IF STILL IN DIFFICULTY COME BACK.
IF CENTRE IS NOT ORIGIN BUT H,K ,THEN THE EQN OF HYPERBOLA WILL BE
((X-H)^2/A^2)-((Y-K)^2/B^2)=1
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Find the equations of the vertical and horizontal asymptotes for the graph of the rational function whose equation is f(x) = x/x+3.
LET Y =X/(X+3)..D.R IS ZERO AT X=-3..SO THIS IS A CRITICAL POINT WHERE THE FUNCTION IS NOT DEFINED.
HENCE WE SPLIT THE DOMAIN OF X
1. FROM -INFINITY TO LESSTHAN -3
2. AND GREATER THAN -3 TO +INFINITY
THE GRAPH FOR DOMAIN
1. FROM -INFINITY TO LESSTHAN -3 IS AS FOLLOWS.

ALGEBRAICALLY , WE FIND THE RANGE OF Y VARIES FROM 1 TO INFINITY.
SO ASYMPTOTES ARE Y=1 AS X TENDS TO MINUS INFINITY
AND Y TENDING TO INFINITY AS X APPROACHES -3
THE GRAPH FOR DOMAIN
2. FROM GREATER THAN -3 TO +INFINITY IS AS FOLLOW

ALGEBRAICALLY , WE FIND THE RANGE OF Y VARIES FROM MINUS INFINITY TO 1.
SO ASYMPTOTES ARE Y=1 AS X TENDS TO INFINITY
AND Y TENDING TO MINUS INFINITY AS X APPROACHES -3