- GIVEN THE STANDARD EQUATION OF A HYPERBOLA, WE WANT TO GRAPH
AND FIND THE MAJOR COMPONENTS OF THE HYPERBOLA.
THE FIRST THING WE NEED TO RECOGNIZE IS THAT
THE FORM OF THE EQUATION TELLS US WHETHER WE HAVE A HORIZONTAL
OR A VERTICAL TRANSVERSE AXIS.
NOTICE IN OUR EQUATION WE HAVE
THE X PART OF THE EQUATION MINUS THE Y PART.
SO BECAUSE THE X PART IS FIRST,
OR THE X PART IS THE POSITIVE FRACTION,
WE'LL HAVE A HORIZONTAL TRANSVERSE AXIS,
SOMETIMES CALLED THE MAJOR AXIS.
AND THAT WOULD BE THIS AXIS HERE.
JUST TO REVIEW, IF THE Y PART OF THE EQUATION WAS FIRST,
AND THEN WE HAD MINUS THE X PART,
WE'D HAVE A VERTICAL TRANSVERSE AXIS,
AND THEREFORE THE HYPERBOLA WOULD OPEN UP AND DOWN.
BUT OUR EQUATION FITS THIS FORM,
SO THE HYPERBOLA WILL OPEN LEFT AND RIGHT.
NEXT, THE CENTER OF THE HYPERBOLA
WILL HAVE COORDINATES H,K.
SO BECAUSE WE HAVE THE QUANTITY X - 2 SQUARED,
THE X COORDINATE IS GOING TO BE 2.
AND BECAUSE WE HAVE THE QUANTITY Y + 1 SQUARED,
THE Y COORDINATE WILL BE -1.
SO OUR CENTER HAS COORDINATES 2,-1,
WHICH WOULD BE THIS POINT HERE.
NEXT, THE DENOMINATOR OF THE FIRST FRACTION,
OR WHAT I THINK OF AS THE POSITIVE FRACTION,
IS GOING TO BE "A" SQUARED.
SO "A" SQUARED = 9.
THEN THE DENOMINATOR OF THE FRACTION
THAT WE'RE SUBTRACTING IS B SQUARED, SO B SQUARED = 4.
SO WE HAVE "A" SQUARED = 9,
SO THE POSITIVE VALUE OF "A" WOULD BE "A" = 3.
AND B SQUARED = 4, SO THE POSITIVE VALUE OF B
WOULD BE B = 2.
KNOWING THESE TWO VALUES HELP US
FIND THE VERTICES OF THE HYPERBOLA,
AS WELL AS THE END POINTS OF THE CONJUGATE AXIS.
THE VERTICES ARE ALSO THE END POINTS
OF THE TRANSVERSE OR MAJOR AXIS,
AND BECAUSE WE HAVE A HORIZONTAL TRANSVERSE AXIS, THE END POINTS,
OR THE VERTICES, WILL BE "A" UNITS TO THE LEFT
AND RIGHT OF THE CENTER.
AND SINCE "A" = 3, ONE VERTEX WILL BE 3 UNITS TO THE LEFT,
OR THIS POINT HERE,
THE OTHER VERTEX WOULD BE 3 UNITS TO THE RIGHT,
OR THIS POINT HERE.
LOOKING AT THE COORDINATE PLANE
WE CAN SEE THE COORDINATES OF THE TWO VERTICES
WOULD BE -1,-1 AND 5,-1.
WE COULD'VE ALSO FOUND THESE BY ADDING AND SUBTRACTING 3
TO THE X COORDINATE OF THE CENTER.
NOTICE 2 - 3 = -1, 2 + 3 = 5.
NOW LET'S COME BACK TO THE FOCI.
LET'S FIND THE END POINTS OF THE CONJUGATE AXIS.
WELL, THIS IS THE CONJUGATE AXIS THAT HAS A LENGTH OF 2 x B.
AND SINCE THE CENTER IS THE MIDPOINT,
FROM THE CENTER IF WE GO UP AND DOWN B UNITS,
WE'LL FIND THE TWO END POINTS OF THE CONJUGATE AXIS,
WHICH IS ALSO CALLED THE MINOR AXIS.
AND SINCE B = 2, IF WE GO DOWN 2 UNITS
THIS WOULD BE ONE END POINT.
IF WE GO UP 2 UNITS,
THIS WOULD BE THE OTHER END POINT OF THE CONJUGATE AXIS.
SO AGAIN, LOOKING AT THE COORDINATE PLANE
WE CAN SEE THE END POINTS WOULD HAVE COORDINATES 2,-3 AND 2,1.
OF COURSE WE COULD ALSO ADD OR SUBTRACT 2
FROM THE Y COORDINATE OF THE CENTER
TO OBTAIN THESE COORDINATES.
NOTICE THAT -1 - 2 = -3, -1 + 2 = 1.
NOW THAT WE HAVE THE END POINTS
OF THE TRANSVERSE AND CONJUGATE AXIS,
WE WANT TO FORM A RECTANGLE THAT WILL HELP US SKETCH
THE ASYMPTOTES OF THE HYPERBOLA.
WE'RE GOING TO FORM A RECTANGLE SO THAT THESE FOUR BLACK POINTS
WILL BE THE MIDPOINTS OF EACH SIDE,
SO THE RECTANGLE WOULD LOOK LIKE THIS.
AGAIN, THIS IS THE TRANSVERSE AXIS, OR MAJOR AXIS,
AND THIS WOULD BE THE CONJUGATE, OR MINOR AXIS.
AND THE ASYMPTOTES OF THE HYPERBOLA
WOULD CONTAIN THE DIAGONALS OF THIS RECTANGLE,
SO OUR TWO ASYMPTOTES WOULD LOOK LIKE THIS AND LOOK LIKE THIS.
SO BEFORE WE FIND THE EQUATION OF THE ASYMPTOTES THOUGH,
LET'S GO AHEAD AND FIND THE COORDINATES OF THE FOCI.
AND AGAIN, BECAUSE WE HAVE A HORIZONTAL MAJOR AXIS,
THE TWO FOCI WILL BE C UNITS
TO THE LEFT AND RIGHT OF THE CENTER.
NOW, WE DON'T HAVE THE VALUE OF C.
WE'LL HAVE TO FIND IT USING
THE EQUATION C SQUARED = "A" SQUARED + B SQUARED,
SINCE WE KNOW "A" SQUARED AND B SQUARED.
SO WE WOULD HAVE C SQUARED = "A" SQUARED, OR 9,
+ B SQUARED, OR 4.
SO C SQUARED = 13.
WE'RE ONLY CONCERNED ABOUT THE POSITIVE VALUE OF C,
SO WE'LL TAKE THE PRINCIPLE SQUARE ROOT.
THIS ISN'T GOING TO SIMPLIFY.
WE'RE GOING TO HAVE C = SQUARE ROOT 13.
BUT BECAUSE WE'RE GOING TO USE THIS VALUE
TO PLOT POINTS ON THE COORDINATE PLANE,
WE ALSO WANT THE DECIMAL APPROXIMATION.
THIS IS APPROXIMATELY 3.606.
SO GOING BACK TO OUR GRAPH,
THE TWO FOCI WILL BE APPROXIMATELY 3.6 UNITS
TO THE RIGHT AND LEFT OF THE CENTER.
SO STARTING AT THE CENTER, IF WE GO TO THE LEFT
APPROXIMATELY 3.6 UNITS IT WOULD BE ABOUT HERE,
AND 3.6 UNITS TO THE RIGHT OF THE CENTER
WOULD BE APPROXIMATELY HERE.
NOTICE HOW TO FIND THE EXACT COORDINATES OF THE FOCI
WE WOULD HAVE TO ADD AND SUBTRACT
SQUARE ROOT OF 13 TO THE X COORDINATE OF THE CENTER,
BECAUSE THE FOCI ARE TO THE LEFT AND RIGHT OF THE CENTER.
SO BECAUSE THE CENTER HAS COORDINATES 2,-1,
THE FOCUS ON THE LEFT WOULD HAVE COORDINATES
2 - SQUARE ROOT 13,-1,
AND THE FOCUS ON THE RIGHT WOULD HAVE COORDINATES
2 + SQUARE ROOT 13,-1.
AND NOW FOR THE LAST STEP WE'RE GOING TO FIND
THE EQUATIONS OF THE ASYMPTOTES,
AND SINCE I'M NOT A BIG FAN OF MEMORIZING FORMULAS,
WE'LL GO AHEAD AND USE THE GRAPHS OF OUR LINES
TO FIND OUR EQUATIONS.
FOR EXAMPLE, IF WE CONSIDER THE ASYMPTOTES,
OR THESE TWO LINES HERE,
NOTICE HOW BOTH LINES CONTAIN THE CENTER,
OR CONTAIN THE POINT 2,-1.
SO IF WE CAN FIND THE SLOPE OF THESE LINES,
WE CAN THEN USE POINT SLOPE FORM OF A LINE TO FIND THE EQUATION.
AND THIS IS ANOTHER REASON WHY THIS RECTANGLE IS VERY HELPFUL.
FOR EXAMPLE, IF WE CONSIDER AGAIN THE CENTER
AND THIS POINT HERE, WE CAN DETERMINE THE SLOPE.
WE WOULD GO UP B UNITS AND RIGHT "A" UNITS,
OR UP 2 UNITS AND RIGHT 3 UNITS.
THEREFORE THE SLOPE OF THIS LINE, WE'LL CALL IT M SUB 1,
IS EQUAL TO 2/3.
THEN LOOKING AT THE OTHER LINE,
NOTICE HOW WE'D HAVE TO GO DOWN B UNITS AND RIGHT "A" UNITS,
OR DOWN 2 UNITS AND RIGHT 3 UNITS.
SO THE SLOPE OF THE SECOND ASYMPTOTE, M SUB 2, = -2/3.
SO NOW WE KNOW THE SLOPES OF THE LINES,
AND WE KNOW THAT THE LINES BOTH CONTAIN THE POINT 2,-1,
WE CAN FIND THE EQUATIONS OF OUR ASYMPTOTES.
LET'S GO AHEAD AND DO THIS ON THE NEXT SLIDE.
SO NOW USING POINT SLOPE FORM OF A LINE,
WHERE THE POINT WOULD BE THE CENTER OF THE HYPERBOLA
THAT BOTH LINES CONTAIN,
AND WE JUST FOUND THE SLOPES OF THE TWO ASYMPTOTES,
ONE EQUATION OF THE ASYMPTOTE WOULD BE
Y - -1 OR Y + 1 = THE SLOPE 2/3 x THE QUANTITY X - X SUB 1
OR X - 2.
WHICH MEANS THE SECOND EQUATION WOULD BE
Y + 1 = -2/3 x THE QUANTITY X - 2.
LET'S GO AHEAD AND CLEAR THE PARENTHESES AND SOLVE FOR Y
SO OUR EQUATIONS ARE IN SLOPED INTERCEPT FORM.
SO WE'LL HAVE Y + 1 = 2/3X.
2/3 x -2 = -4/3.
NOW WE'LL GO AHEAD AND SUBTRACT 1 ON BOTH SIDES.
SUBTRACTING 1 IS THE SAME AS SUBTRACTING 3/3.
-4/3 - 3/3 = -7/3.
NOW FOR THE SECOND LINE WE'D HAVE Y + 1 = -2/3X + 4/3.
AGAIN, SUBTRACT 1 ON BOTH SIDES.
SO WE HAVE Y = -2/3X.
THIS WOULD BE 4/3 - 3/3 OR 1/3.
THESE WOULD BE THE TWO EQUATIONS OF OUR ASYMPTOTES.
AND NOW LET'S FINISH BY GRAPHING OUR HYPERBOLA.
WE KNOW THE HYPERBOLA WILL CONTAIN EACH VERTEX
AND APPROACH THE ASYMPTOTES.
SO THE GRAPH OF THE HYPERBOLA
WOULD LOOK SOMETHING LIKE THIS, AND LIKE THIS.
OKAY, HOPE YOU FOUND THIS EXPLANATION HELPFUL.