- WELCOME TO A SECOND VIDEO ON THE HYPERBOLA.
THE GOALS OF THIS VIDEO ARE TO CONVERT AN EQUATION
OF A HYPERBOLA IN GENERAL FORM TO STANDARD FORM,
AND THEN ALSO TO GRAPH THE HYPERBOLA.
LET'S GO AHEAD AND DO A QUICK REVIEW FROM THE LAST VIDEO.
IF WE HAVE THE EQUATION OF A HYPERBOLA IN STANDARD FORM
AND THE X PART OF THE EQUATION IS A POSITIVE PART,
THE HYPERBOLA WILL HAVE A HORIZONTAL TRANSVERSE AXIS.
IF THE Y PART OF THE EQUATION IS THE POSITIVE PART,
THEN THE TRANSVERSE AXIS WILL BE VERTICAL
AS WE SEE HERE ON THE RIGHT.
NEXT, THE CENTER OF THE HYPERBOLA IN BOTH CASES
WILL BE (H,K).
A SQUARED WILL BE THE DENOMINATOR
OF THE POSITIVE PART OF THE EQUATION.
AND B SQUARED WILL BE THE DENOMINATOR OF THE PART
THAT WE'RE SUBTRACTING.
OR YOU CAN THINK OF IT AS THE NEGATIVE PART.
ONCE WE DETERMINE THE VALUE OF A AND B
WE CAN DETERMINE THE VALUE OF C
BY USING THE EQUATION C SQUARED
= A SQUARED + B SQUARED.
SO ONCE WE DETERMINE WHAT TYPE OF TRANSVERSE AXIS
THE HYPERBOLA HAS THEN THE CENTER,
AND THEN A, B, AND C,
WE CAN MAKE A NICE GRAPH OF THE HYPERBOLA.
AND HERE IS HOW WE DO IT.
WE FIRST START BY PLOTTING THE CENTER IN BOTH CASES.
NEXT, IF IT HAS A HORIZONTAL TRANSVERSE AXIS,
WE'LL ADD AND SUBTRACT A TO THE X COORDINATE
OF THE CENTER TO DETERMINE THE TWO VERTICES OF THE HYPERBOLA
OR THESE TWO POINTS ON THE HYPERBOLA.
IF IT HAS A VERTICAL TRANSVERSE AXIS
WE'LL ADD AND SUBTRACT A TO THE Y COORDINATE OF THE CENTER
AS WE SEE HERE AND HERE.
THE NEXT THING WE'RE GOING TO DO IS CONSTRUCT THIS RECTANGLE
THAT IS CENTERED BETWEEN THE TWO PIECES OF THE HYPERBOLA.
AND THIS RECTANGLE HAS DIMENSIONS 2A x 2B.
SO FROM THE CENTER WE'LL GO UP AND DOWN B UNITS
SO THAT WE HAVE THESE FOUR POINTS ON THE RECTANGLE,
AND THEN WE'LL FORM THE RECTANGLE.
THE REASON THIS RECTANGLE IS SO HELPFUL
IS THAT THE LINES PASSING THROUGH THE DIAGONALS
HERE AND HERE ARE THE ASYMPTOTES TO THE HYPERBOLA.
THEREFORE THESE TWO PIECES OF THE HYPERBOLA
WILL APPROACH THOSE TWO LINES.
AGAIN, IT'S VERY SIMILAR
WHEN WE HAVE A VERTICAL TRANSVERSE AXIS,
EXCEPT NOW WE'LL ADD AND SUBTRACT B FROM THE CENTER
TO GET THIS POINT AND THIS POINT,
AND THEN FORM THE RECTANGLE.
AND, AGAIN, WE'LL CONSTRUCT THE ASYMPTOTES
TO AIDE US IN MAKING A NICE, ACCURATE GRAPH.
THE LAST THING WE'LL DO IS DETERMINE THE COORDINATES
OF THE FOCI.
IF IT HAS A HORIZONTAL TRANSVERSE AXIS,
WE WILL ADD AND SUBTRACT C TO THE X COORDINATE
AT THE CENTER.
AND IF IT HAS A VERTICAL TRANSVERSE AXIS,
WE'LL ADD AND SUBTRACT C FROM THE Y COORDINATE
OF THE CENTER.
LET'S GO AHEAD AND START WITH AN EQUATION IN GENERAL FORM,
WRITE IT IN STANDARD FORM AND THEN GRAPH IT.
NORMALLY, IN ORDER TO WRITE AN EQUATION IN STANDARD FORM
WE HAVE TO COMPLETE THE SQUARE.
HOWEVER, ON THIS EQUATION,
SINCE WE ONLY HAVE AN X SQUARED TERM
AND Y SQUARED TERM,
WE DON'T HAVE TO DO THAT.
WE CAN JUST DIVIDE BY 16.
SO THIS WILL GIVE US X SQUARED OVER FOUR
MINUS Y SQUARED OVER 16 EQUALS ONE.
AND NOW THIS IS IN STANDARD FORM.
NOW, IF WE REALLY WANTED TO WE COULD REWRITE THIS SLIGHTLY
SO IT MATCHES THIS FORM EXACTLY.
WHAT I MEAN BY THAT IS WE COULD WRITE THIS
AS THE QUANTITY X - 0 SQUARED
OVER 4 - Y - 0 SQUARED OVER 16 = 1.
AND THIS JUST REINFORCES
THAT OUR CENTER IS GOING TO BE THE ORIGIN (0,0).
NEXT, SINCE THE X PART OF THE EQUATION IS THE POSITIVE PART,
THIS HYPERBOLA WILL HAVE A HORIZONTAL TRANSVERSE AXIS.
SO IT'S GOING TO LOOK SOMETHING LIKE THIS
WHEN WE FINISH,
OPEN TO THE RIGHT AND OPEN TO THE LEFT.
THIS IS IMPORTANT BECAUSE IF WE KNOW THIS,
WE SHOULDN'T HAVE TO MEMORIZE ALL OF THESE RULES
IN ORDER TO DETERMINE THE VERTICES AND THE FOCI.
A SQUARED IS GOING TO EQUAL FOUR
BECAUSE THAT IS THE DENOMINATOR
OF THE POSITIVE PART OF THE EQUATION.
THEREFORE A = 2.
B SQUARED WILL EQUAL 16.
SO B = 4.
NOW, LET'S GO AHEAD AND FIND C.
REMEMBER C SQUARED = A SQUARED + B SQUARED.
SO YOU WOULD HAVE C SQUARED = 4 + 16.
THAT'S 20.
SO C = SQUARE ROOT OF 20,
WHICH IS EQUAL TO TWO SQUARED OF FIVE.
AND SINCE WE USED C TO DETERMINE THE FOCI,
AND WE'RE GOING TO BE PLOTTING THOSE,
LET'S GO AHEAD AND CONVERT C TO A DECIMAL.
I'VE ALREADY DONE THAT.
IT'S APPROXIMATELY 4.47.
LET'S GO AHEAD AND TAKE ALL OF THIS INFORMATION
OVER TO THE NEXT SCREEN AND THEN WE CAN MAKE A GRAPH.
OKAY. SO WE HAVE THE CENTER AND THE VALUE OF A, B AND C.
AND WE KNOW WE HAVE A HORIZONTAL TRANSVERSE AXIS.
SO WE'LL START BY PLOTTING THE CENTER, (0,0).
NEXT, WE'RE GOING TO ADD AND SUBTRACT A
FROM THE X COORDINATE OF THE CENTER
TO DETERMINE THE TWO VERTICES
WHICH ARE TWO POINTS ON THE HYPERBOLA.
SINCE A = 2, THIS WOULD BE THE POINT (2,0),
AND THIS WOULD BE THE POINT (-2,0)
AND THESE ARE OUR VERTICES.
NEXT, WE'LL CONSTRUCT THIS RECTANGLE
SO WE CAN SKETCH IN THE ASYMPTOTES
TO HELP US MAKE A NICE GRAPH.
AND NOTICE FROM HERE WE WENT LEFT AND RIGHT A UNITS.
NOW, WE'RE GOING TO GO UP AND DOWN B UNITS.
SINCE B = 4, WE'LL GO UP TO FOUR AND DOWN TO -4.
WE'RE GOING TO USE THE TWO GREEN POINTS
AND THE TWO VERTICES TO CONSTRUCT THIS RECTANGLE.
IT WILL LOOK SOMETHING LIKE THIS.
NOW WE'LL SKETCH THE ASYMPTOTES
WHICH WILL PASS THROUGH THE DIAGONALS OF THE RECTANGLE.
SO FROM HERE WE CAN ACTUALLY MAKE A NICE GRAPH
OF THE HYPERBOLA.
IT WILL START HERE AT THIS VERTEX
AND APPROACH THESE TWO ASYMPTOTES.
SO IT WILL LOOK SOMETHING LIKE THIS.
AND THEN ON THE LEFT IT WILL LOOK SOMETHING LIKE THIS.
BUT, HOWEVER, WE'RE ALSO OFTEN ASKED TO FIND THE FOCI.
SO LET'S GO AHEAD AND DO THAT.
REMEMBER, THE FOCI WILL BE SOMEWHERE OVER HERE
ON THE RIGHT,
AND SOMEWHERE OVER HERE ON THE LEFT.
SO WE'LL ADD AND SUBTRACT C
TO THE X COORDINATE OF THE CENTER.
WELL, SINCE THE CENTER IS (0,0),
WE'RE GOING TO HAVE 0 + 4.74 WHICH WOULD JUST BE (4.74,0).
AND THE SECOND FOCUS WOULD BE 0 - 4.47 OR (-4.47,0).
SO ONE OF THEM WILL BE SOMEWHERE IN HERE.
THE OTHER ONE WILL BE SOMEWHERE OVER HERE.
LET'S GO AHEAD AND TAKE A LOOK AT ONE MORE.
SO NOW WE ARE GOING TO HAVE TO COMPLETE THIS SQUARED.
SO LETS FIRST GROUP
THE X TERMS AND THE Y TERMS TOGETHER.
SINCE THE Y SQUARED TERM IS POSITIVE,
I'M GOING TO GO AHEAD AND PUT THAT TERM FIRST.
WE'LL ALSO GO AHEAD AND SUBTRACT 11 ON BOTH SIDES.
NOW, IN ORDER TO COMPLETE THE SQUARE
ON THE Y PART AND THE X PART,
WE WANT THE LEADING COEFFICIENT TO BE ONE.
SO HERE WE'LL FACTOR OUT THE FOUR
AND LEAVE SOME ROOM FOR THE NUMBER.
NOW, HERE WE'RE GOING TO FACTOR OUT A -1.
SO IT WILL BE - +X SQUARED + 2X.
OKAY, NOW TO COMPLETE THE SQUARE WE WILL TAKE HALF OF -4
AND THEN SQUARE IT.
-2 SQUARED WOULD BE FOUR.
BY PUTTING A FOUR HERE WE'RE ACTUALLY ADDING 16.
SO WE'LL HAVE TO ADD 16 ON THE RIGHT.
TO COMPLETE THIS SQUARED HERE WE'LL TAKE HALF OF TWO.
THAT'S ONE AND SQUARE IT.
SO WE'LL ADD ONE HERE.
BECAUSE OF THIS MINUS SIGN HERE,
WE'RE ACTUALLY SUBTRACTING ONE.
SO WE HAVE TO SUBTRACT ONE ON THE RIGHT.
LET'S GO AHEAD AND FACTOR NOW.
AT 4 x (Y - 2) SQUARED - (X + 1) SQUARED = 4.
TO MAKE THE RIGHT SIDE ONE
WE'LL DIVIDE EVERYTHING BY FOUR AND SIMPLIFY.
SO WE'RE GOING TO HAVE (Y - 2) SQUARED
DIVIDED BY 1 - (X + 1) SQUARED DIVIDED BY 4 = 1.
THE FIRST THING WE NOTICE HERE
IS THE Y PART OF THE EQUATION IS POSITIVE.
SO NOW WE'RE GOING TO HAVE A VERTICAL TRANSVERSE AXIS.
SO THE HYPERBOLA WILL LOOK SOMETHING LIKE THIS
WHEN WE GRAPH IT.
THE CENTER IS LOCATED AT (-1,+2).
AND THIS DENOMINATOR HERE WILL BE A SQUARED.
SO A = 1.
THIS IS B SQUARED. SO B = 2.
LET'S GO AHEAD AND FIND C NOW.
C SQUARED = A SQUARED + B SQUARED.
WELL, A SQUARED = 1, AND B SQUARED = 4.
SO WE HAVE C SQUARED = 5.
SO C = SQUARE ROOT OF 5,
WHICH IS APPROXIMATELY EQUAL TO 2.24.
OKAY, ONCE AGAIN, LET'S GO AHEAD
AND TAKE THIS INFORMATION TO THE NEXT SCREEN.
SO LET'S GO AHEAD AND START BY PLOTTING OUR CENTER
AT (-1,2) RIGHT HERE.
FROM HERE,
SINCE WE KNOW WE HAVE A VERTICAL TRANSVERSE AXIS,
WE'RE GOING TO ADD AND SUBTRACT A
TO THE Y COORDINATE OF THE CENTER
TO DETERMINE THE TWO VERTICES.
SINCE A = 1 WE'LL GO UP ONE UNIT AND DOWN ONE UNIT
FROM THE CENTER TO GET TWO VERTICES.
SO THIS WILL BE (-1,3) AND THIS WILL BE (-1,1).
NEXT, SINCE WE'VE WENT UP AND DOWN A UNITS
TO DETERMINE THE TWO VERTICES,
NOW WE'LL GO LEFT AND RIGHT B UNITS
TO DETERMINE THAT RECTANGLE TO SKETCH IN THE ASYMPTOTES.
SINCE B = 2, WE'LL GO TWO UNITS TO THE RIGHT
AND TWO UNITS TO THE LEFT.
AND WE'RE GOING TO USE THESE GREEN POINTS
AND THESE RED POINTS TO PERFORM A RECTANGLE LIKE SO.
AND THE LINES PASSING THROUGH THE DIAGONALS
WILL BE THE ASYMPTOTES.
SO FROM HERE WE CAN MAKE A NICE GRAPH OF THE HYPERBOLA.
IT STARTS AT THIS VERTEX AND APPROACHES THE TWO ASYMPTOTES
IN THE SAME BELOW.
AND WE SHOULD END BY FINDING THE COORDINATES
OF THE TWO FOCI.
SINCE THE FOCI WILL BE UP HERE AND DOWN HERE,
WE WILL ADD AND SUBTRACT C
TO THE Y COORDINATE OF THE CENTER.
LET'S GO AHEAD AND WRITE THAT OUT.
SO WE'LL HAVE (-1,4.24)
AND THE SECOND FOCUS WOULD BE (-1,-.24).
SOMEWHERE IN HERE IS WHERE THE SECOND POINT IS.
OKAY. THAT WILL DO IT FOR THIS VIDEO.