Tip:
Highlight text to annotate it
X
- WELCOME TO THE SECOND VIDEO ON THE CONIC SECTIONS THE ELLIPSE.
THE GOAL OF THIS VIDEO IS TO GRAPH AN ELLIPSE
GIVEN IN GENERAL FORM.
BUT BEFORE WE DO THAT LETS GO AHEAD AND REVIEW.
IF WE HAVE THE EQUATION OF AN ELLIPSE IN STANDARD FORM,
THE CENTER WILL BE (H, K).
SINCE WE KNOW THAT "A" IS GREATER THAN B,
THE FRACTION WITH THE LARGER DENOMINATOR TELLS YOU WHETHER
THE MAJOR AXIS WILL BE HORIZONTAL OR VERTICAL.
IF THE LARGER DENOMINATOR IS UNDER THE X PART OF THE EQUATION
THEN WE WILL HAVE A HORIZONTAL MAJOR AXIS.
IF THE LARGER DENOMINATOR
IS UNDER THE Y PART OF THE EQUATION,
WE WILL HAVE A VERTICAL AXIS.
IF WE CAN DETERMINE THE VALUE OF "A",
"A" WILL TELL US THE DISTANCE
FROM THE CENTER TO THE ENDPOINTS OF THE MAJOR AXIS
AND B WILL TELL US THE DISTANCE
FROM THE CENTER TO THE ENDPOINTS OF THE MINOR AXIS.
AND THEN LASTLY, WE CAN USE THIS EQUATION HERE
TO DETERMINE THE VALUE OF C,
AND C WILL BE THE DISTANCE FROM THE CENTER TO THE TWO FOCI.
AND NOTICE ALL THE INFORMATION IS THE SAME
WHETHER WE HAVE A HORIZONTAL OR VERTICAL MAJOR AXIS
EXCEPT TO FIND THE FOCI.
WHEN THERE'S A HORIZONTAL MAJOR AXIS YOU ADD AND SUBTRACT C
FROM THE X COORDINATE OF THE CENTER.
WHILE IF WE HAVE A VERTICAL MAJOR AXIS
WE ADD OR SUBTRACT C FROM THE Y COORDINATES OF THE CENTER
TO DETERMINE THE COORDINATES OF THE FOCI.
SO THE GOAL OF THIS VIDEO IS TO BE ABLE
TO WRITE A PARABOLA GIVEN IN GENERAL FORM,
IN STANDARD FORM SO WE CAN GRAPH IT.
SO THE FIRST THING WE'RE GOING TO DO
IS GROUP THE X TERMS AND THE Y TERMS TOGETHER
AND MOVE THE CONSTANT TO THE RIGHT SIDE.
SO WE'D HAVE X SQUARED
THAT'S THE ONLY X TERM + THEN
WE WOULD HAVE 4Y SQUARED - 16Y = IT WOULD BE -12.
NOW WE DON'T HAVE TO COMPLETE THE SQUARE ON THE X PART
BECAUSE IT'S ALREADY A PERFECT SQUARE.
WE CAN REWRITE THIS AS X - 0 SQUARED IF WE WANT.
NOW TO COMPLETE THE SQUARE ON THE Y PART
WE'RE GOING TO HAVE TO HAVE A COEFFICIENT OF 1.
SO WE'RE GOING TO FACTOR OUT THE 4
AND WE'LL BE LEFT WITH Y SQUARED THIS WILL BE - 4Y
IF WE FACTOR OUT THE 4 AND WE'LL LEAVE ROOM
TO COMPLETE THE SQUARE AND THIS WILL EQUAL -12.
SO TO COMPLETE THE SQUARE HERE
WE'RE GOING TO TAKE HALF OF -4 AND THEN SQUARE IT.
WELL -2 SQUARED WOULD BE 4.
NOW WE HAVE TO BE CAREFUL HERE.
WHEN WE ADD A 4 HERE, WE'RE ACTUALLY ADDING 16
BECAUSE WHEN WE DISTRIBUTE WE'D HAVE A 4 x 4.
SO WE HAVE TO ADD 16 TO THE RIGHT SIDE OF THE EQUATION
TO MAINTAIN THE EQUALITY.
LET'S GO AHEAD AND REWRITE THIS NOW.
WE'D HAVE THE QUANTITY
X - 0 SQUARED + 4 x THE QUANTITY Y - 2 SQUARED.
IT'S NOW A PERFECT SQUARE = 4.
NOW FOR THE EQUATION TO BE IN STANDARD FORM IT MUST EQUAL 1.
SO NOW WE'LL DIVIDE EVERYTHING BY 4.
SO OUR EQUATION IN STANDARD FORM WILL BE THE QUANTITY
X - 0 SQUARED DIVIDED BY 4 PLUS
NOTICE HERE THE 4 AND THE 4 SIMPLIFY OUT
SO WE'D HAVE Y - 2 SQUARED DIVIDED BY 1 = 1.
OKAY, SO NOW WE SHOULD BE ABLE TO DETERMINE
A LOT OF INFORMATION ABOUT THE GRAPH OF THIS ELLIPSE.
FIRST, THE CENTER WILL BE (0, 2).
NOTICE THE LARGER DENOMINATOR IS UNDER THE X PART
SO WE'RE GOING TO HAVE A HORIZONTAL MAJOR AXIS
AS PICTURED HERE.
SO NOW WE KNOW THAT "A" SQUARED = 4,
SO "A" = 2 AND B SQUARED = 1 SO B = 1.
WHILE WE'RE HERE LET'S GO AHEAD AND FIND THE VALUE OF C.
REMEMBER "A" IS THE LARGEST
SO THE EQUATION WOULD BE "A" SQUARED = B SQUARED + C SQUARED.
SO "A" SQUARED IS 4, B SQUARED IS 1.
SO WE'D HAVE C SQUARED = 3.
SO C = THE SQUARE ROOT OF 3.
SO NOW WE HAVE ALL THE INFORMATION WE NEED
TO MAKE A NICE GRAPH OF THIS ELLIPSE.
LET'S GO AHEAD AND TRANSFER THIS INFORMATION
ON THE NEXT SCREEN AND GRAPH IT.
THE CENTER WAS (0, 2), "A" = 2, B = 1,
AND C = THE SQUARE ROOT OF 3
AND WE ALSO KNOW WE HAVE A HORIZONTAL MAJOR AXIS.
LET'S GO AHEAD AND PLOT OUR CENTER.
THIS IS (0, 2).
SO TO FIND THE ENDPOINTS OF THE MAJOR AXIS
WE'LL GO RIGHT 2 UNITS AND THEN LEFT 2 UNITS OF THE CENTER.
AND THEN THE ENDPOINTS OF THE MINOR AXIS
WILL BE B UNITS ABOVE THE CENTER AND B UNITS BELOW THE CENTER.
SO IT WOULD BE HERE AND HERE.
LET'S GO AHEAD AND SKETCH OUR ELLIPSE.
NEXT, WE SHOULD PLOT THE FOCI.
REMEMBER IF THIS IS THE CENTER,
WE KNOW THE FOCI WILL BE OVER HERE SOMEWHERE
AND OVER HERE SOMEWHERE.
SO WE'RE GOING TO ADD AND SUBTRACT
THE SQUARE ROOT OF 3 FROM THE X COORDINATE OF THE CENTER.
SO ONE OF THE FOCI WILL BE WELL 0 + (SQUARE ROOT OF 3, 2)
AND THE OTHER FOCI WILL BE
(0 - THE SQUARE ROOT 3 OR -SQUARE ROOT, 2)
LETS GO AHEAD AND CONVERT SQUARE ROOT 3 TO A DECIMAL
SO WE CAN PLOT IT.
SO IT'S APPROXIMATELY 1.73.
SO (1.73, 2) IS SOMEWHERE OVER HERE
AND (-1.73, 2) WOULD BE SOMEWHERE OVER HERE.
THESE ARE OUR 2 FOCI AND THIS WAS F1,
THIS WAS F2 BASED UPON HOW WE LABELED IT HERE.
LET'S GO AHEAD AND TRY ANOTHER ONE.
SO THE FIRST STEP IS TO GROUP THE X TERMS
AND THE Y TERMS TOGETHER.
SO WE'LL HAVE 9X SQUARED - 18X + 4Y SQUARED + 16Y = 11.
NEXT STEP, TO COMPLETE THE SQUARE
WE DO HAVE TO FACTOR OUT THE LEADING COEFFICIENTS
FOR THE X PART AND THE Y PART.
SO WE'LL HAVE 9 x THE QUANTITY X SQUARED - 2X + SOME NUMBER
+ 4 x THE QUANTITY Y SQUARED + 4Y + SOME NUMBER MUST EQUAL 11.
OKAY, SO WE'LL TAKE HALF OF -2 AND SQUARE IT.
THAT WILL GIVE US + 1.
NOW IF WE PUT A 1 HERE IT'S REALLY LIKE ADDING 9.
SO WE'LL ADD 9 TO THE RIGHT SIDE.
NOW HERE WE'LL TAKE HALF OF 4 AND SQUARE IT.
THAT WILL GIVE US 4.
PUTTING A 4 HERE IS LIKE ADDING 16.
SO WE HAVE TO ADD 16 TO THE RIGHT SIDE.
THESE SHOULD NOW BE PERFECT SQUARE TRINOMIALS
LETS GO AHEAD AND TRY TO FACTOR IT.
HERE WE'LL HAVE THE QUANTITY
X - 1 SQUARED + 4 x THE FACTORS OF 4 THAT ADD TO 4
THAT WOULD BE 2 AND 2.
SO WE'D HAVE THE QUANTITY X + 2 SQUARED MUST EQUAL.
OVER HERE, WE'RE GOING TO HAVE 11 + 9 THAT'S 20 + 16, 36.
NOW TO MAKE IT EQUAL TO 1
WE'RE GOING TO HAVE TO DIVIDE EVERYTHING BY 36.
IF WE SIMPLIFY THESE FRACTIONS, IT SHOULD BE IN STANDARD FORM.
LET'S SEE. WELL 9/36 THAT WOULD BE 1/4.
SO THE DENOMINATOR IS 4 PLUS
HERE WE'RE GOING TO HAVE THE QUANTITY Y + 2 SQUARED.
THIS WOULD BE 1/9.
SO OUR DENOMINATOR IS 9 = 1.
NOTICE THE LARGER DENOMINATOR IS UNDER THE Y PART THIS TIME.
SO NOW WE'RE GOING TO HAVE A VERTICAL MAJOR AXIS
AS WE SEE HERE.
OUR CENTER WOULD HAVE THE COORDINATES (1, -2).
"A" SQUARED WOULD EQUAL 9.
THAT IMPLIES "A" = 3 AND B SQUARED = 4.
SO WE HAVE B = 2.
LET'S TAKE THIS INFORMATION OVER TO THE NEXT SCREEN.
WE STILL HAVE TO FIND C AND THEN WE'LL MAKE A GRAPH.
OKAY, SO WE HAVE OUR CENTER AND WE HAVE "A" AND B.
REMEMBER THE EQUATION TO FIND
C IS "A" SQUARED = B SQUARED + C SQUARED.
SO WE KNOW "A" SQUARED IS 9, WE KNOW B SQUARED IS 4.
SO IT LOOKS LIKE C SQUARED = 5.
SO C = THE SQUARE ROOT OF 5.
OKAY, LET'S SEE IF WE CAN MAKE A NICE GRAPH OF THIS NOW.
LET'S FIRST PLOT OUR CENTER, (1, -2) WOULD BE HERE.
NEXT, WE KNOW WE HAVE A VERTICAL MAJOR AXIS
SO TO FIND THE ENDPOINTS OF THE MAJOR AXIS
WILL GO UP "A" UNITS AND DOWN "A" UNITS.
"A" = 3.
SO WE'LL GO UP 3 UNITS, DOWN 3 UNITS.
B IS EQUAL TO 2 SO WE'LL GO RIGHT FROM THE CENTER TWO UNITS
AND LEFT FROM THE CENTER TWO UNITS.
SO OUR ELLIPSE PASSES THROUGH THESE 4 POINTS.
THE ONLY THING LEFT TO DO NOW IS TO DETERMINE
THE COORDINATES OF THE FOCI.
NOW REMEMBER SINCE THE TWO FOCI WILL BE UP HERE AND DOWN HERE,
WE'RE GOING TO ADD OR SUBTRACT
SQUARE ROOT 5 FROM THE Y COORDINATE OF THE CENTER.
SO THE FIRST FOCUS WOULD BE (1, -2 + THE SQUARE ROOT OF 5)
AND THE SECOND FOCUS WOULD BE (1, -2 - SQUARE ROOT 5).
AGAIN, IT'S ALWAYS HELPFUL TO CONVERT THESE TO DECIMALS
TO GRAPH THEM ON THE COORDINATE PLANE.
SO LET'S GO AHEAD AND DO THAT.
SO ONE OF THEM WOULD BE RIGHT ABOUT HERE
AND THE OTHER FOCUS WOULD BE SOMEWHERE AROUND HERE.
OKAY, THAT'S GOING TO DO IT FOR THIS VIDEO.
I HOPE YOU FOUND THESE EXAMPLES HELPFUL.
THANK YOU FOR WATCHING.