SOLUTION: byperbolas, last one i swear. first i would appreciate it if u defined some words for me and tell me what they are/do. Asymptote, transverse axis,what the a^2 and b^2 do. Next I

Question 27832: byperbolas, last one i swear.
first i would appreciate it if u defined some words for me and tell me what they are/do. Asymptote, transverse axis,what the a^2 and b^2 do. Next I dont get how to graph the hyperbolas, more specific how to pull info from the graph to the equation, whats standard form for a hyperbola. and last section my book gave me a formula xy=c is a hypebola with the x and y axis as asymptotes(dont know what it means) Sketch the graph of each hyperbola. I promis I'll leave ya alone just lay some stuff out for me please!!
thanks!
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 .
****************************************************************************
GIVEN BELOW ARE SOME MORE EXAMPLES.NOW YOU TRY TO DRAW YOUR REQUIRED CURVE AND UNDERSTAND.IF STILL IN DIFFICULTY COME BACK.
******************************************************************************
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

RELATED QUESTIONS
Hello could you please help me with these questions, I would really appreciate it if you... (answered by rothauserc,MathTherapy)

hi there, I asked the following question before and I had a solution given to me, I do... (answered by stanbon,bucky)

okay hi the section iam doing is graphing linear Inequalities. and now iam doing it says... (answered by elima)

Hi, My name is Lauren and I really need help with my homework. Can you please help me... (answered by stanbon)

Hello, I appeciate very much that there are people like you up there; willin and... (answered by ewatrrr)

Can anyone help me? I swear it's the last one:-)My problem is: Use the... (answered by Alan3354)

i need sum help on my math we are learning inequalities right now with greater than and... (answered by stanbon)

I am brushing up on my math skills for the school year and have come upon a problem that... (answered by Alan3354)

One-fourth of the cars purchased at a dealership are luxury models. If 360 luxury models... (answered by richwmiller,wilft1)