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- WELCOME TO ANOTHER EXAMPLE ON HOW TO GRAPH AND FIND
THE MAJOR COMPONENTS OF A HYPERBOLA WHEN THE EQUATION
IS GIVEN IN STANDARD FORM.
THE FIRST THING WE NEED TO RECOGNIZE IS THAT
THE FORM OF THE EQUATION TELLS US WHETHER
WE'LL HAVE A VERTICAL OR A HORIZONTAL TRANSVERSE AXIS.
LOOKING AT OUR EQUATION,
SINCE WE HAVE THE Y PART OF THE EQUATION - THE X PART
OF THE EQUATION, WE ARE GOING TO HAVE A VERTICAL TRANSVERSE AXIS,
WHICH MEANS THE HYPERBOLA WILL OPEN UP AND DOWN.
SO I ALWAYS THINK OF THIS AS IF THE POSITIVE FRACTION
IS THE Y PART, WE HAVE A VERTICAL TRANSVERSE AXIS.
IF THE X PART OF THE EQUATION IS POSITIVE,
THEN WE HAVE A HORIZONTAL TRANSVERSE AXIS.
SO TO REVIEW THIS AGAIN, IF THE X PART OF THE EQUATION
IS FIRST - THE Y PART,
WE'LL HAVE A HORIZONTAL TRANSVERSE AXIS.
SO THE HYPERBOLA OPENS LEFT AND RIGHT.
IF THE Y PART OF THE EQUATION IS FIRST - THE X PART,
WE HAVE A VERTICAL TRANSVERSE AXIS,
AND THE HYPERBOLA OPENS UP AND DOWN.
NOW THAT WE KNOW WE HAVE A VERTICAL TRANSVERSE AXIS,
LET'S GO AHEAD AND FIND THE CENTER OF THE HYPERBOLA.
IT'LL HAVE COORDINATES H,K.
AND SINCE WE HAVE THE QUANTITY X + 2, THE X COORDINATE IS -2.
AND BECAUSE WE HAVE THE QUANTITY Y - 3, THE Y COORDINATE IS 3.
SO -2,3 IS OUR CENTER.
NEXT, THE DENOMINATOR OF THE FIRST FRACTION,
OR THE POSITIVE FRACTION, IN OUR EQUATION = "A" SQUARED,
AND THE DENOMINATOR OF THE FRACTION
WE'RE SUBTRACTING = B SQUARED.
SO WE KNOW THAT "A" SQUARED = 9, AND B SQUARED = 16.
SO IF "A" SQUARED = 9,
THE POSITIVE VALUE OF "A" WOULD BE 3, AND IF B SQUARED = 16,
THE POSITIVE VALUE OF B = 4.
SO KNOWING THE VALUE OF "A" AND B ALLOWS US TO FIND
THE VERTICES, WHICH ARE THE TWO END POINTS OF THE TRANSVERSE,
OR MAJOR AXIS AS WELL AS THE END POINTS OF THE CONJUGATE AXIS,
SOMETIMES CALLED THE MINOR AXIS.
BECAUSE WE HAVE A VERTICAL TRANSVERSE AXIS,
THE TWO VERTICES WILL BE "A" UNITS
ABOVE AND BELOW THE CENTER.
AND SINCE "A" = 3, THIS IS THE CENTER,
ONE VERTEX WOULD BE HERE 3 UNITS ABOVE THE CENTER,
THE OTHER VERTEX WOULD BE 3 UNITS BELOW THE CENTER, OR HERE.
SO NOTICE THE COORDINATES WOULD BE -2,0 AND -2,6.
AGAIN, THESE ARE ALSO THE END POINTS OF THE TRANSVERSE AXIS,
AND NOW WE'LL FIND THE FOCI LATER.
TO FIND THE END POINTS OF THE CONJUGATE AXIS,
THE END POINTS WILL BE B UNITS
TO THE LEFT AND RIGHT OF THE CENTER.
SINCE THIS IS THE CONJUGATE AXIS,
IT HAS A LENGTH OF 2 x B WHERE THE CENTER IS THE MIDPOINT.
SO IF B = 4, ONE END POINT WOULD BE 4 UNITS
TO THE LEFT OF THE CENTER HERE.
THE OTHER END POINT WOULD BE 4 UNITS
TO THE RIGHT OF THE CENTER, WHICH WOULD BE HERE.
AND AGAIN, THIS SEGMENT HERE
WOULD BE THE CONJUGATE AXIS OR MINOR AXIS.
LOOKING AT THE COORDINATE PLANE
WE CAN SEE THE COORDINATES WOULD BE -6,3 AND 2,3.
NOW WE COULD HAVE FOUND THE VERTICES
BY ADDING AND SUBTRACTING 3 TO THE Y COORDINATE OF THE CENTER,
AND WE COULD FIND THE END POINTS OF THE CONJUGATE AXIS
BY ADDING AND SUBTRACTING 4 FROM THE X COORDINATE OF THE CENTER
IF WE WANTED TO.
BUT I THINK DOING IT GRAPHICALLY MAKES A LOT MORE SENSE.
NOW, FROM HERE WE'RE GOING TO SKETCH A RECTANGLE
THAT'LL HELP US SKETCH THE ASYMPTOTES OF THE HYPERBOLA,
WHERE THE FOUR END POINTS THAT WE HAVE HERE
WILL BE THE MIDPOINT OF EACH SIDE OF THE RECTANGLE.
SO THE RECTANGLE WOULD LOOK LIKE THIS.
NOTICE HOW THE LENGTH OF THESE TWO SIDES WOULD BE 2A.
THE LENGTH OF THESE TWO SIDES WOULD BE 2B.
AND THE ASYMPTOTES OF THE HYPERBOLA
WOULD CONTAIN THE DIAGONALS OF THIS RECTANGLE.
SO ONE ASYMPTOTE WOULD BE THIS LINE HERE,
AND THE OTHER ASYMPTOTE WOULD BE THIS LINE HERE.
AND HAVING THESE GRAPHED ON THE COORDINATE PLANE
WILL MAKE IT MUCH EASIER FOR US
TO FIND THE EQUATION OF THESE ASYMPTOTES,
BUT BEFORE WE DO THIS,
LET'S FIND THE COORDINATES OF THE TWO FOCI.
AND AGAIN, BECAUSE WE HAVE A VERTICAL MAJOR AXIS,
THE TWO FOCI WILL BE C UNITS ABOVE AND BELOW THE CENTER.
WE DON'T HAVE THE VALUE OF C,
BUT WE CAN FIND C USING THE EQUATION
C SQUARED = "A" SQUARED + B SQUARED,
WHICH IS ALWAYS TRUE FOR A HYPERBOLA.
REMEMBER, WE ALREADY KNOW "A" SQUARED AND B SQUARED.
SO USING OUR EQUATION AND THE KNOWN VALUES
OF "A" SQUARED AND B SQUARED,
WHICH MEANS C SQUARED MUST EQUAL "A" SQUARED, OR 9, + B SQUARED,
WHICH IS 16.
C SQUARED = 25.
WE'RE ONLY CONCERNED ABOUT THE POSITIVE VALUE OF C,
SO WE'LL TAKE THE PRINCIPLE SQUARE ROOT OF BOTH SIDES.
THE SQUARE ROOT OF 25 IS 5, SO WE HAVE C = 5.
WELL, IF C = 5, THAT MEANS THE TWO FOCI
ARE 5 UNITS ABOVE AND BELOW THE CENTER.
SO ONE FOCUS WOULD BE HERE,
AND THE OTHER FOCUS WOULD BE 5 UNITS BELOW THE CENTER, OR HERE,
WHICH MEANS THE COORDINATES WOULD BE -2,8 AND -2,-2.
WE COULD ALSO FIND THESE COORDINATES
BY ADDING AND SUBTRACTING 5 FROM THE Y COORDINATE OF THE CENTER.
NOW, BEFORE WE GRAPH OUR HYPERBOLA,
LET'S FIND THE EQUATIONS OF THE ASYMPTOTES.
LET'S START BY FINDING THE SLOPE OF EACH OF THESE LINES,
AND WE CAN DO THIS USING THE VALUES OF "A" AND B,
OR JUST USING THE COORDINATE PLANE.
LET'S TAKE A LOOK AT THIS LINE FIRST THAT HAS A POSITIVE SLOPE.
TO FIND THE SLOPE, LET'S USE THE CENTER OF THE HYPERBOLA,
THIS POINT HERE AND THIS POINT HERE.
NOTICE IF WE WANT TO MOVE FROM THIS POINT TO THIS POINT
WE HAVE TO GO UP "A" UNITS, OR UP 3 UNITS,
AND RIGHT B UNITS, OR RIGHT 4 UNITS.
SO IF THIS IS 3 AND THIS WOULD BE 4,
THE SLOPE OF THIS LINE IS 3/4.
SO WE'LL CALL THIS M SUB 1.
AND NOW LOOKING AT THE SECOND LINE, OR THIS LINE HERE,
THAT HAS A NEGATIVE SLOPE.
IF WE USE THE CENTER AND THIS POINT HERE,
NOTICE HOW WE'D HAVE TO GO DOWN 3 UNITS, OR DOWN "A" UNITS,
AND RIGHT 4 UNITS, OR RIGHT B UNITS.
SO THIS WOULD ACTUALLY BE -3, AND THIS WOULD BE 4,
SO WE CAN SEE THE SLOPE OF THIS LINE WOULD BE -3/4.
SO M SUB 2 = -3/4.
THIS IS REALLY ALL WE NEED TO FIND THE EQUATIONS
OF THESE TWO LINES,
SINCE WE HAVE THE SLOPES
AND WE ALSO KNOW ONE POINT ON BOTH LINES.
SO WE'LL FIND THE EQUATIONS
USING POINT SLOPE FORM OF A LINE,
AND THEN WE'LL SOLVE THE EQUATION FOR Y.
SO AGAIN, WE'RE GOING TO USE POINT SLOPE FORM OF A LINE.
WE KNOW BOTH LINES PASS THROUGH THE CENTER, OR THIS POINT HERE,
WHERE ONE LINE HAS A SLOPE OF 3/4,
THE OTHER LINE HAS SLOPE -3/4.
WHICH MEANS ONE EQUATION
WOULD BE Y - 3 = THE SLOPE 3/4 x THE QUANTITY X - -2 OR X + 2.
THE SECOND EQUATION WOULD JUST BE
Y - 3 = -3/4 x THE QUANTITY X + 2.
SO THESE ARE THE EQUATIONS OF THE TWO ASYMPTOTES.
LET'S GO AHEAD AND SOLVE FOR Y. SO WE'LL CLEAR THE PARENTHESES,
AND THEN SOLVE FOR Y.
SO WE'LL HAVE Y - 3 = HERE WE'LL DISTRIBUTE, 3/4X.
3/4 x 2 WOULD BE 6/4 OR 3/2, SO + 3/2.
NOW WE'LL GO AND ADD 3 TO BOTH SIDES.
SO WE HAVE Y = 3/4X.
ADDING 3 IS THE SAME AS ADDING 6/2.
3/2 + 6/2 WOULD BE 9/2.
AND THEN FOR THE SECOND LINE,
AGAIN WE'LL FIRST CLEAR THE PARENTHESES.
HERE WE'RE DISTRIBUTING -3/4, SO IT'S -3/4X - 6/4 OR - 3/2.
AND THEN AGAIN WE'LL ADD 3 TO BOTH SIDES,
SO WE HAVE Y = -3/4X, AND THEN -3/2 WOULD BE THE SAME AS + 6/2.
THAT WOULD BE 3/2 OR + 3/2.
SO THESE WOULD BE THE TWO EQUATIONS OF OUR ASYMPTOTES.
NOW LET'S GO AHEAD AND GRAPH OUR HYPERBOLA.
WE KNOW IT'S GOING TO CONTAIN THE VERTICES
AND APPROACH THE ASYMPTOTES.
SO THE BRANCH ON TOP WOULD LOOK SOMETHING LIKE THIS,
AND THE BRANCH ON THE BOTTOM WOULD LOOK SOMETHING LIKE THIS.
OKAY, I HOPE YOU FOUND THIS EXPLANATION HELPFUL.