- IN THIS EXAMPLE WE'RE GIVEN THE GENERAL EQUATION
OF AN ELLIPSE AND ASKED TO WRITE THE EQUATION IN STANDARD FORM,
AS WELL AS GRAPH AND FIND THE KEY COMPONENTS OF THE ELLIPSE.
THE STANDARD EQUATION OF AN ELLIPSE IS GIVEN HERE.
SO IN ORDER TO WRITE THIS IN STANDARD FORM,
WE'LL HAVE TO FORM A PERFECT SQUARE TRINOMIAL
IN TERMS OF X AND Y TO OBTAIN THE QUANTITY X - H SQUARED
AND THE QUANTITY Y - K SQUARED.
SO FOR THE FIRST STEP
WE'RE GOING TO GROUP THE X TERMS AND Y TERMS TOGETHER
AND MOVE THE CONSTANT TO THE RIGHT SIDE.
SO WE'D HAVE 4X SQUARED AND THEN - 32X,
THEN + 9Y SQUARED - 54Y.
THEN WE'RE GOING TO SUBTRACT 109 ON BOTH SIDES,
SO WE WOULD HAVE = -109.
AND NOW WE'RE GOING TO COMPLETE
THIS SQUARE ON THE X PART AND Y PART OF THE EQUATION,
BUT BEFORE WE DO THIS WE WANT THE COEFFICIENT
OF THE X SQUARED TERM
AND THE COEFFICIENT OF THE Y SQUARED TERM TO BE 1.
SO WE'RE GOING TO FACTOR OUT 4 FROM THESE FIRST TWO TERMS.
SO WE'LL HAVE 4 x THE QUANTITY X SQUARED.
TO FACTOR OUT 4 FROM 32 THAT WOULD BE 8, SO WE'LL - 8X.
AND NOW WE'RE GOING TO FORM OUR PERFECT SQUARE TRINOMIAL
INSIDE THE PARENTHESIS.
SO WE'LL HAVE PLUS A CONSTANT, WHICH WE'LL FIND IN A MINUTE.
PLUS NOW WE'LL FACTOR OUT 9 FROM THE SECOND TWO TERMS.
SO WE'LL HAVE 9 X THE QUANTITY Y SQUARED - 6Y.
AND, AGAIN, WE'RE GOING TO COMPLETE THIS SQUARE
INSIDE THE PARENTHESIS,
SO WE'LL HAVE TO ADD A CONSTANT HERE.
THIS IS GOING TO BE = -109 + TWO CONSTANTS
TO MAINTAIN EQUALITY WITH THE LEFT SIDE.
AND TO COMPLETE THIS SQUARE ON X SQUARED - 8X
WE'RE GOING TO TAKE HALF OF - 8,
WHICH WOULD BE -4 AND THEN SQUARE IT, WHICH WOULD BE 16.
SO WE'RE GOING TO ADD 16 HERE,
WHICH WILL MAKE THIS A PERFECT SQUARE TRINOMIAL.
BUT BECAUSE OF THIS 4 OUT HERE,
WE'RE ACTUALLY ADDING 4 x 16 TO THE LEFT SIDE,
WHICH WOULD BE 64.
SO TO MAINTAIN EQUALITY WE'LL HAVE TO ADD 64
TO THE RIGHT SIDE.
SO WE CAN'T FORGET BECAUSE THIS 4 HERE,
WE'RE NOT JUST ADDING 16, WE'RE ADDING 64.
AND NOW TO COMPLETE THIS SQUARE ON Y SQUARED - 6Y
WE'LL TAKE HALF OF -6 AND SQUARE IT.
THAT'S -3 SQUARED THAT'S 9.
SO WE'LL ADD 9 HERE, BUT NOTICE IF WE DISTRIBUTE,
WE'RE ACTUALLY ADDING 81,
SO WE'LL HAVE TO ADD 81 TO THE RIGHT SIDE TO MAINTAIN EQUALITY.
NOW WE'LL GO AHEAD AND FACTOR THIS TRINOMIAL
AND THIS TRINOMIAL,
WHICH, AGAIN, SHOULD BE PERFECT SQUARED TRINOMIALS.
SO WE'LL HAVE 4 X TWO BINOMIALS WHERE WE HAVE X AND X.
THE FACTORS OF 16 THAT ADD TO -8 WOULD BE -4 AND -4.
NOTICE HOW WE DO HAVE A PERFECT SQUARE TRINOMIAL
BECAUSE OUR TWO FACTORS ARE EQUAL + 9
AND THEN WE HAVE Y AND Y,
THE FACTORS OF 9 THAT ADD TO -6 WOULD BE -3 AND -3.
AGAIN, NOTICE HOW WE HAVE TWO EQUAL FACTORS.
AND THEN -109 + 64 + 81 = 36.
SO NOW WE'RE GETTING CLOSE.
WE HAVE 4 x THE QUANTITY
X - 4 SQUARED + 9 x THE QUANTITY Y - 3 SQUARED = 36.
REMEMBER OUR EQUATION MUST = 1.
SO BECAUSE WE HAVE 36 ON THE RIGHT SIDE,
WE'LL NOW DIVIDE EVERYTHING BY 36.
WE'LL DIVIDE THIS BY 36, THIS BY 36, AND THIS BY 36.
AND THESE WILL SIMPLIFY NICELY.
THERE'S ONE 4 IN 4 AND NINE 4's IN 36.
THERE'S ONE 9 IN 9, AND FOUR 9's IN 36,
AND THIS SIMPLIFIES NICELY TO 1/1.
SO OUR STANDARD EQUATION IS THE QUANTITY X - 4 SQUARED
DIVIDED BY 9 + THE QUANTITY Y - 3 SQUARED DIVIDED BY 4 = 1.
NOTICE HOW THE LARGER DENOMINATOR
IS UNDER THE X PART OF THE EQUATION,
THEREFORE OUR ELLIPSE WILL HAVE A HORIZONTAL MAJOR AXIS.
SO NOW WE'LL GO AHEAD AND FIND THE KEY COMPONENTS
AND GRAPH THIS ELLIPSE.
SO, AGAIN, BECAUSE THE LARGER DENOMINATOR
IS UNDER THE Y PART OF THE EQUATION
WE KNOW WE'RE GOING TO HAVE A HORIZONTAL MAJOR AXIS.
SO OUR ELLIPSE WILL LOOK SOMETHING LIKE THIS
WHERE THE CENTER HAS COORDINATES (H,K).
THE DISTANCE FROM THE CENTER TO THE TWO ENDPOINTS
OF THE MAJOR AXIS OR THE VERTICES IS EQUAL TO "A" UNITS.
AND THE DISTANCE FROM THE CENTER TO THE TWO ENDPOINTS
OF THE MINOR AXIS IS EQUAL TO B UNITS.
SO LOOKING AT OUR EQUATION,
BECAUSE WE HAVE X - 4 THE X COORDINATE OF THE CENTER
WOULD BE 4,
AND BECAUSE WE HAVE Y - 3 THE Y COORDINATE OF THE CENTER
WOULD BE 3.
SO LET'S GO AHEAD AND PLOT THIS CENTER,
WHICH WOULD BE HERE.
NEXT BECAUSE THE LARGER DENOMINATOR IS 9
WE KNOW "A" SQUARED = 9.
BECAUSE WE'RE ONLY CONCERNED ABOUT THE POSITIVE VALUE OF "A",
WE KNOW THAT "A" = 3.
THE SMALLER DENOMINATOR WOULD BE B SQUARED,
SO B SQUARED = 4.
AND THE POSITIVE VALUE OF B WOULD BE B = 2.
SO NOW WE KNOW THE VALUES OF A AND B,
WE CAN FIND THE ENDPOINTS OF THE MAJOR AXIS,
ALSO CALLED THE VERTICES,
AND THE ENDPOINTS OF THE MINOR AXIS.
SO BECAUSE WE KNOW WE HAVE A HORIZONTAL MAJOR AXIS,
THE ENDPOINTS OF THE MAJOR AXIS
WILL BE "A" UNITS TO THE LEFT AND RIGHT OF THE CENTER.
SO IF THIS IS OUR CENTER, AND "A" = 3,
ONE ENDPOINT OF THE MAJOR AXIS WOULD BE HERE 3 UNITS
TO THE RIGHT.
THE OTHER ENDPOINT WOULD BE 3 UNITS TO THE LEFT OR HERE.
SO WE CAN LOOK AT THE COORDINATE PLANE
AND TELL THE ENDPOINTS OF THE MAJOR AXIS
OR OUR VERTICES WOULD HAVE COORDINATES (1,3)
AND COORDINATES (7,3).
OF COURSE, WE ALSO COULD'VE ADDED AND SUBTRACTED 3
FROM THE X COORDINATE OF THE CENTER.
SO, AGAIN, THIS IS THE MAJOR AXIS,
AND SINCE B = 2 THIS GIVES US THE DISTANCE
FROM THE CENTER TO THE TWO ENDPOINTS OF THE MINOR AXIS.
SO ONE ENDPOINT OF THE MINOR AXIS WOULD BE 2 UNITS
ABOVE THE CENTER OR HERE,
AND THE OTHER ENDPOINT WOULD BE 2 UNITS BELOW THE CENTER,
WHICH WOULD BE HERE.
SO THE COORDINATES WOULD BE (4,1) AND (4,5).
NOTICE HOW WE ALSO COULD'VE ADDED AND SUBTRACTED 2
FROM THE Y COORDINATE OF THE CENTER TO FIND THESE ENDPOINTS.
NEXT WE'RE ASKED TO FIND THE COORDINATES OF THE FOCI.
GOING BACK TO OUR DIAGRAM HERE,
NOTICE HOW THE FOCI WILL BE C UNITS
TO THE LEFT AND RIGHT OF THE CENTER,
AGAIN, SINCE WE HAVE HORIZONTAL MAJOR AXIS.
WE DON'T KNOW THE VALUE OF C YET,
BUT WE DO KNOW A SQUARED AND B SQUARED,
THEREFORE WE CAN USE THE EQUATION
"A" SQUARED = B SQUARED + C SQUARED TO FIND C SQUARED,
AND THAT'S C.
LET'S START BY SOLVING THIS EQUATION FOR C SQUARED
BY SUBTRACTING B SQUARED ON BOTH SIDES.
THAT WOULD GIVE US C SQUARED = A SQUARED - B SQUARED.
SO WE WOULD HAVE C SQUARED = A SQUARED,
WHICH IS 9 - B SQUARED, WHICH IS 4.
SO C SQUARED = 5.
WE'RE ONLY CONCERNED ABOUT THE POSITIVE VALUE OF C,
SO WE'LL TAKE THE PRINCIPAL SQUARE ROOT OF BOTH SIDES.
SO C = SQUARE ROOT 5.
BECAUSE WE'RE GOING TO USE THIS TO PLOT POINTS
ON THE COORDINATE PLANE,
LETS GET A DECIMAL APPROXIMATION.
THIS SHOULD BE APPROXIMATELY 2.236.
SO GOING BACK TO OUR GRAPH,
WE NOW KNOW C = SQUARE ROOT 5 OR APPROXIMATELY 2.236,
WHICH MEANS THE TWO FOCI WILL BE APPROXIMATELY 2.236 UNITS
TO THE LEFT AND RIGHT OF THE CENTER,
SO APPROXIMATELY HERE AND HERE.
NOTICE HOW WE WOULD HAVE TO ADD AND SUBTRACT C
TO THE X COORDINATE OF THE CENTER
TO FIND THE COORDINATES OF THE FOCI.
SO IF THE CENTER HAS COORDINATES (4,3),
THE FOCUS ON THE LEFT WOULD HAVE COORDINATES
(4 - SQUARE ROOT 5,3).
AND THE POINT ON THE RIGHT WOULD HAVE COORDINATES
(4 + SQUARE ROOT 5,3).
NOW WE WANT TO FIND THE ECCENTRICITY,
WHICH = C DIVIDED BY A.
WHERE WE KNOW C IS SQUARE ROOT 5 AND A = 3.
DECIMAL VALUE OF THIS IS APPROXIMATELY 0.745.
THE CLOSER THIS VALUE IS TO 1 THE MORE ELONGATED
THE ELLIPSE IS.
THE CLOSER IT IS TO 0 THE MORE CIRCULAR THE ELLIPSE IS.
AND FINALLY WE'LL GRAPH OUR ELLIPSE.
THE GRAPH WILL LOOK SOMETHING LIKE THIS.
OKAY. I HOPE YOU FOUND THIS HELPFUL.