SOLUTION: It's the sick one again, Im in a conic section and it wants me to write and equation for the hyperbola described: the vertecies of the hyperbola are at (0,-4) and (0.4). the Conju

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: It's the sick one again, Im in a conic section and it wants me to write and equation for the hyperbola described: the vertecies of the hyperbola are at (0,-4) and (0.4). the Conju      Log On


   

Question 27829: It's the sick one again,
Im in a conic section and it wants me to write and equation for the hyperbola described: the vertecies of the hyperbola are at (0,-4) and (0.4). the Conjugate,(define it for me please) axis has a length of 14 units. Please explain this to me it wont help if i dont get it.
-Sick and Desperate
Answer by venugopalramana(3286) About Me  (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.
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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.
+graph%28+500%2C+500%2C+-50%2C1%2C+-1%2C+50%2C+x%2F%28x%2B3%29%2C1%29+
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
+graph%28+500%2C+500%2C+-4%2C+50%2C+-50%2C+2%2C+x%2F%28x%2B3%29+%2C1%29+
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