- WELCOME TO THE FIRST OF TWO PARTS ON THE CONIC SECTION,
THE HYPERBOLA.
THE GOAL OF THIS VIDEO IS TO GRAPH THE HYPERBOLA
IN STANDARD FORM.
A HYPERBOLA IS A SET OF ALL POINTS (X,Y)
FOR WHICH THE ABSOLUTE VALUE OF THE DIFFERENCES
BETWEEN THE DISTANCES FROM TWO DISTINCT FIXED POINTS,
CALLED FOCI, IS CONSTANT.
NOW, I THINK THE BEST WAY TO GET A FEEL FOR THIS DEFINITION
IS TO TAKE A LOOK AT AN APPLET.
THE HYPERBOLA IS GRAPHED HERE IN RED.
THE TWO FOCI ARE HERE AND HERE.
SO WHAT YOU WILL SEE IS TWO SEGMENTS FROM THIS POINT
IN BLUE TO BOTH OF THE FOCI,
AND WHAT THAT DEFINITION IS SAYING
IS IF WE FIND THE DIFFERENCE OF THESE TWO LENGTHS
AND THEN TAKE THE OPPOSITE VALUE OF THAT,
IT WILL BE THE SAME FOR ANY POINT ON THE HYPERBOLA.
SO, FOR EXAMPLE, RIGHT NOW,
I'M NOT SURE IF YOU COULD SEE IT,
BUT THEY'RE TAKING THIS SHORT DISTANCE HERE
AND SUBTRACTING THIS LONGER DISTANCE.
AND THAT DIFFERENCE IS -8.
SO IF WE TAKE THE OPPOSITE VALUE OF THAT IT WOULD BE 8.
AND THAT WILL BE TRUE FOR ANY POINT ON THE HYPERBOLA.
IF I MOVE THIS POINT, AS I ANIMATE THIS POINT,
THE COMMON DIFFERENCE REMAINS -8,
AND THE OPPOSITE VALUE OF THAT WOULD BE 8.
IF I MOVE THIS POINT TO THE OTHER PIECE OF THE HYPERBOLA,
THE COMMON DIFFERENCE IS NOW 8.
AND, AGAIN, THE OPPOSITE VALUE OF 8 WILL STILL BE EIGHT.
LET'S FIRST TAKE A LOOK AT SOME OF THE KEY FEATURES
OF THE GRAPH OF A HYPERBOLA.
THE FIRST THING YOU ARE GOING TO PROBABLY NOTICE
IS THAT HYPERBOLAS CAN OPEN LEFT AND RIGHT OR UP AND DOWN.
AND THAT IS BASED UPON WHETHER THE TRANSVERSE AXIS
IS HORIZONTAL AS WE SEE HERE.
THIS IS THE TRANSVERSE AXIS.
OR WHETHER THE TRANSVERSE AXIS IS VERTICAL AS WE SEE HERE.
HYPERBOLAS HAVE A CENTER AS WE SEE HERE IN BLUE.
HYPERBOLAS ALSO HAVE TWO POINTS
THAT ARE ON THE HYPERBOLA END ON THE TRANSVERSE AXIS
CALLED VERTICES AS WE SEE HERE IN RED AND HERE IN RED.
ALSO ON THE TRANSVERSE AXIS ARE THE FOCI.
AND YOU CAN THINK OF THOSE AS BEING INSIDE THE HYPERBOLA.
AGAIN, WHETHER THE TRANSVERSE AXIS IS HORIZONTAL
OR WHETHER IT IS VERTICAL.
AND THEN LASTLY YOU WILL SEE THESE GREEN LINES
THAT ARE CALLED ASYMPTOTES.
THESE ARE LINES THAT THEY HYPERBOLA APPROACHES,
BUT IT WILL NOT CROSS,
AND THESE ARE VERY HELPFUL WHEN WE GO TO MAKE THE GRAPH.
SO LETS TAKE A LOOK AT
HOW WE'RE GOING TO BE ABLE TO DETERMINE
ALL OF THESE KEY COMPONENTS
WHEN THE GRAPH IS IN STANDARD FORM.
NOW, THIS IS A BUSY DIAGRAM,
BUT IT CONTAINS EVERYTHING YOU NEED TO KNOW
IN ORDER TO GRAPH THE HYPERBOLA IN STANDARD FORM.
FIRST, WE'LL NOTICE THIS LOOKS VERY SIMILAR TO AN ELLIPSE,
BUT NOW WE HAVE A DIFFERENCE RATHER THAN A SUM.
AND WHEN THE X PART OF THE EQUATION IS THE POSITIVE PART,
NOTICE THAT THE HYPERBOLA WILL OPEN LEFT AND RIGHT
AND WILL HAVE A HORIZONTAL TRANSVERSE AXIS.
IF THE Y PART OF THE EQUATION IS POSITIVE,
THEN THE TRANSVERSE AXIS IS VERTICAL,
AND THE HYPERBOLA WILL OPEN UP AND OPEN DOWN.
NOTICE THAT A SQUARED WILL ALWAYS BE
UNDER THE POSITIVE PART OF THE EQUATION,
AND B SQUARED WILL BE UNDER THE DIFFERENCE
OR THE NEGATIVE PART OF THE EQUATION.
ONCE WE IDENTIFY A SQUARED AND B SQUARED
WE CAN FIND C SQUARED
BY USING THE EQUATION C SQUARED
= A SQUARED + B SQUARED
BECAUSE NOW C WILL BE THE LONGEST LENGTH.
ONCE WE FIND A, B AND C WE START BY LOCATING THE CENTER
WHICH HAS THE COORDINATES (H,K)
AND IF WE HAVE A HORIZONTAL TRANSVERSE AXIS
WE'LL ADD OR SUBTRACT A TO THE X COORDINATE
TO FIND THE COORDINATES OF THE VERTICES
AND WE'LL ADD AND SUBTRACT C TO THE X COORDINATE.
AND WE'LL ADD OR SUBTRACT C TO THE X COORDINATE OF THE CENTER
TO DETERMINE THE COORDINATES OF THE FOCI.
NOW, WHAT I WANT YOU TO NOTICE IS THIS RECTANGLE
THAT'S BETWEEN THE TWO PIECES OF HYPERBOLA.
IT HAS THE DIMENSIONS 2A x 2B.
AND THE REASON THIS RECTANGLE IS SO HELPFUL
IS THAT LINES PASSING THROUGH THE DIAGONALS
OF THIS RECTANGLE WILL BE THE ASYMPTOTES OF THE HYPERBOLA
AND THEREFORE ARE VERY HELPFUL IN MAKING THE GRAPH.
EVERYTHING IS VERY SIMILAR
WHEN WE HAVE A VERTICAL TRANSVERSE AXIS
EXCEPT NOW WE'LL BE ADDING AND SUBTRACTING A AND C
TO THE Y COORDINATE OF THE CENTER
TO DETERMINE THE VERTICES AND THE FOCI.
LET'S GO AHEAD AND LOOK AT AN EXAMPLE
AND SEE IF WE CAN PUT ALL OF THIS TOGETHER.
THE FIRST THING YOU SHOULD NOTICE ON THIS EQUATION
IS THAT THE X PART IS THE POSITIVE PART
WHICH MEANS WE'LL HAVE A HORIZONTAL TRANSVERSE AXIS.
SO OUR GRAPH SHOULD LOOK SOMETHING LIKE THIS
AS WE SEE DOWN BELOW.
NEXT, THE CENTER OF THE HYPERBOLA
WILL HAVE THE COORDINATES (H,K).
SO THE X COORDINATE WILL BE 2,
AND THE Y COORDINATE WILL BE -3.
A SQUARED WILL BE UNDER THE POSITIVE PART OF THE EQUATION.
SO A SQUARED = 4.
SO WE KNOW A = 2.
AND B SQUARED = 9.
SO B = 3.
LET'S GO AHEAD AND FIND THE VALUE OF C.
REMEMBER, C SQUARED = A SQUARED + B SQUARED.
SO C SQUARED WILL = 4 + 9.
THAT'S 13.
SO WE'LL HAVE C = SQUARE ROOT OF 13.
AND I ALREADY FOUND THE DECIMAL APPROXIMATION FOR THIS
BECAUSE THAT'S HELPFUL WHEN WE'RE GRAPHING.
SO C IS APPROXIMATELY EQUAL TO 3.61.
OKAY, SO LET'S GO AHEAD
AND START BY PLOTTING THE CENTER AT (2,-3).
AGAIN, KEEPING IN MIND THE HYPERBOLA
WILL OPEN TO THE LEFT AND OPEN TO THE RIGHT.
THE TWO VERTICES OF THE HYPERBOLA
WILL BE EIGHT UNITS TO THE RIGHT
AND EIGHT UNITS TO THE LEFT OF THE CENTER.
SO SINCE A = 2, ONE VERTEX WILL BE HERE AT (4,-3),
AND THE OTHER ONE WILL BE AT (0,-3).
REMEMBER, THESE ARE TWO POINTS ON THE HYPERBOLA
BECAUSE WE KNOW IT OPENS SOMETHING LIKE THIS
AND SOMETHING LIKE THIS.
THE NEXT THING WE WANT TO DO IS CREATE THIS RECTANGLE HERE
WITH THE DIMENSIONS 2A x 2B
SO WE CAN SKETCH THOSE ASYMPTOTES.
THE LENGTH FROM THIS RED POINT TO THIS RED POINT IS 2A.
THIS WOULD BE THE WIDTH OF THE RECTANGLE.
THE QUESTION IS, WHAT WOULD THE HEIGHT BE?
WELL, THE HEIGHT WOULD BE 2B.
AND SINCE B = 3, 2B WOULD BE SIX.
THE RECTANGLE WOULD HAVE A HEIGHT OF SIX UNITS.
SO WE'LL SKETCH A RECTANGLE SIX UNITS HIGH
BY FOUR UNITS WIDE LIKE SO.
AND, AGAIN, THE DIAGONALS OF THIS RECTANGLE
WILL BE THE ASYMPTOTES OF THE HYPERBOLA.
LET'S GO AHEAD AND SKETCH THOSE.
BELIEVE IT OR NOT THIS IS ENOUGH INFORMATION
TO MAKE A NICE SKETCH OF THIS HYPERBOLA
BECAUSE WE KNOW EACH PIECE IS GOING TO APPROACH
THESE ASYMPTOTES.
SO IT'S GOING TO LOOK SOMETHING LIKE THIS,
AND ALSO SOMETHING LIKE THIS.
AND WE SHOULD GO AHEAD AND FIND THE COORDINATES
OF THE FOCI.
REMEMBER, THAT'S GOING TO BE C UNITS
TO THE RIGHT AND LEFT OF THE CENTER
WHERE C IS APPROXIMATELY 3.61.
FOCI ONE WOULD BE 2 + 3.61 COMA -3,
AND THE SECOND FOCI WOULD BE 2 - 3.61 COMA -3.
WELL, THIS WOULD BE (5.61,-3) WAY OVER HERE ON THE END.
NOW, THIS FOCI ONE,
AND THIS IS GOING TO END UP BEING (-1.61,-3)
SOMEWHERE IN HERE FOR FOCI TWO.
AND WE'RE NOT GOING TO TAKE THE TIME
TO FIND THE EQUATION OF THESE ASYMPTOTES.
SOMETIMES WE'RE ASKED TO DO THAT,
BUT YOU COULD QUICKLY FIND THE SLOPE OF THESE LINES
AND THEN USE POINT SLOPE FORM TO FIND THE EQUATION.
LET'S GO AHEAD AND TAKE A LOOK AT ONE MORE EXAMPLE.
NOTICE ON THIS HYPERBOLA THE Y PART IS THE POSITIVE PART,
WHICH MEANS OUR TRANSVERSE AXIS IS NOW VERTICAL,
AND THEREFORE THE HYPERBOLA WILL OPEN UP AND OPEN DOWN
AS WE SEE PICTURED HERE.
OUR CENTER WILL HAVE THE COORDINATES (+2,-4).
A SQUARED = 4.
THEREFORE A = 2.
B SQUARED = 16. B = 4.
REMEMBER, A SQUARED IS ALWAYS THE DENOMINATOR
OF THE POSITIVE PART.
AND LET'S GO AHEAD AND FIND C WHILE WE'RE HERE.
C SQUARED = A SQUARED + B SQUARED,
WHICH WOULD BE 4 + 16.
THAT'S 20.
SO C = SQUARE ROOT OF 20 WHICH IS TWO SQUARE ROOT OF FIVE.
THIS IS APPROXIMATELY 4.47.
LET'S GO AHEAD AND SEE IF WE CAN SKETCH THIS.
WE'LL FIRST PLOT THE CENTER, (2,-4).
WE KNOW THIS PARABOLA IS GOING TO OPEN UP AND DOWN.
SO NOW WE'LL HAVE TO ADD A TO THE Y COORDINATE
TO FIND THE TWO VERTICES OF THE HYPERBOLA.
SINCE A = 2 WE'LL GO UP TWO UNITS AND DOWN TWO UNITS.
SO THIS WOULD BE (2,-2) AND THIS WOULD BE (2,-6).
THOSE ARE TWO POINTS ON THE HYPERBOLA.
SO NOW WE'LL GO AHEAD AND SKETCH THE RECTANGLE
SO WE CAN FORM THE ASYMPTOTES.
WE ALREADY KNOW THE HEIGHT OF THE RECTANGLE NOW.
NOW WE NEED TO DETERMINE HOW WIDE IT WOULD BE.
IT WOULD BE 2B UNITS WIDE.
SINCE B = 4 IT IS GOING TO BE EIGHT UNITS WIDE.
SO WE'RE GOING TO SKETCH A RECTANGLE
THAT FROM HERE WOULD BE FOUR UNITS TO THE RIGHT OUT TO SIX,
AND FOUR UNITS TO THE LEFT OUT TO A -2.
FROM HERE WE CAN FINISH OUR RECTANGLE.
IT WOULD GO UP TO THIS HEIGHT AND THEN OVER,
AND THEN BACK DOWN.
NOW, THE ASYMPTOTES WOULD BE FORMED BY THE DIAGONALS
OF THE RECTANGLE HERE AND HERE.
SO FROM THIS VERTEX
THE HYPERBOLA WILL APPROACH THE ASYMPTOTES TO THE LEFT
AND TO THE RIGHT, AND THE SAME BELOW.
BUT WE SHOULD GO AHEAD AND FIND THE COORDINATES
OF THE FOCI.
SO NOW WE WILL ADD AND SUBTRACT C
FROM THE Y COORDINATE OR THE CENTER.
THE FIRST FOCUS OF (2,-4) + 4.47,
AND THE SECOND FOCUS WILL HAVE THE COORDINATES (2,-4) - 4.47.
SO THIS FOCUS WILL BE (2,+.47) SOMEWHERE IN HERE,
AND THIS WILL BE (2,-8.47)
SOMEWHERE DOWN HERE OFF OF THE SCREEN SLIGHTLY.
OKAY. THAT WILL DO IT FOR THIS.