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- WELCOME TO AN EXAMPLE ON HOW TO GRAPH THE STANDARD EQUATION
OF AN ELLIPSE WHEN THE CENTER OF THE ELLIPSE
IS NOT AT THE ORIGIN.
THE FIRST THING WE SHOULD RECOGNIZE
ABOUT THE GIVEN EQUATION IS THAT THE LARGER DENOMINATOR,
IN THIS CASE 25, IS UNDER THE Y PART OF THE EQUATION,
AND THEREFORE THE ELLIPSE WILL HAVE A VERTICAL MAJOR AXIS
AS WE SEE PICTURED HERE.
SO BECAUSE THE LARGER DENOMINATOR
IS UNDER THE Y PART OF THE EQUATION,
OUR EQUATION IS IN THIS FORM HERE WHERE "A" SQUARED
IS A LARGER DENOMINATOR
AND B SQUARED IS THE SMALLER DENOMINATOR.
SO IN THIS CASE WE KNOW "A" SQUARED = 25.
WE'LL BE USING THE POSITIVE VALUE OF "A"
TO HELP US GRAPH OUR ELLIPSE.
AND SINCE 5 SQUARED = 25 WE KNOW "A" = 5.
THE SMALLER DENOMINATOR = B SQUARED,
SO IN THIS CASE B SQUARED = 4.
WE'LL BE USING THE POSITIVE VALUE OF B
TO HELP US GRAPH THE ELLIPSE AS WELL.
SO IF B SQUARED = 4, THE POSITIVE VALUE OF B WOULD BE 2.
NEXT, TO FIND THE CENTER OF THE ELLIPSE,
THE CENTER HAS COORDINATES (H, K).
SO LOOKING AT OUR EQUATION,
AND SINCE WE HAVE THE QUANTITY X - 1,
THE X COORDINATE OF THE CENTER IS 1.
AND BECAUSE WE HAVE THE QUANTITY Y + 4
THE Y COORDINATE OF THE CENTER IS -4.
SO LET'S GO AHEAD AND PLOT THE CENTER OF OUR ELLIPSE,
(1, -4) WOULD BE HERE.
NOW WE WANT TO FIND THE END POINTS OF THE MAJOR AXIS,
ALSO CALLED THE VERTICES OF THE ELLIPSE,
AND ALSO THE END POINTS OF THE MINOR AXIS.
SO LOOKING AT OUR NOTES OVER HERE FOR A MOMENT,
BECAUSE WE HAVE A VERTICAL MAJOR AXIS,
THE TWO END POINTS OF THE MAJOR AXIS OR LONGER AXIS
ARE A UNITS ABOVE AND BELOW THE CENTER.
AND SINCE WE ALREADY KNOW "A" = 5,
FROM THE CENTER IF WE GO UP FIVE UNITS,
WHICH WOULD BE HERE AND DOWN FIVE UNITS FROM THE CENTER,
WHICH WOULD BE HERE.
THESE ARE THE TWO END POINTS OF THE MAJOR AXIS.
SO TO FIND THE COORDINATES OF THESE END POINTS
WE CAN EITHER LOOK AT THE COORDINATE PLANE,
OR ADD AND SUBTRACT FIVE FROM THE Y COORDINATE OF THE CENTER.
THIS POINT HERE WOULD HAVE COORDINATES (1,-9)
AND THIS END POINT WOULD HAVE COORDINATES (1,1).
NOTICE THAT -4 AND -5 IS -9, -4 + 5 IS 1.
AND NOW TO FIND THE END POINTS OF THE MINOR AXIS
OR THE SHORTER AXIS, THIS ONE HERE,
NOTICE HOW THE END POINTS ARE B UNITS
TO THE RIGHT AND LEFT OF THE CENTER.
AND SINCE B = 2,
SO FROM THE CENTER IF WE MOVE 2 UNITS TO THE LEFT HERE
AND 2 UNITS TO THE RIGHT HERE,
THESE WOULD BE THE END POINTS OF THE MINOR AXIS.
SO TO OBTAIN THESE COORDINATES
WE CAN LOOK AT THE COORDINATE PLANE,
OR WE COULD ALSO ADD AND SUBTRACT 2
FROM THE X COORDINATE OF THE CENTER.
NOTICE HOW THIS POINT HERE WOULD HAVE COORDINATES (-1,-4)
AND THIS POINT HERE WOULD HAVE COORDINATES (3,-4).
NOTICE HOW 1 - B OR Y - 2 IS -1, AND 1 + B OR 1 + 2 = 3.
ALSO NOTICE THAT THE LENGTH OF THE MAJOR AXIS IS 2 x A,
AND THE LENGTH OF THE MINOR AXIS IS 2 x B.
NEXT WE WANT TO FIND THE COORDINATES OF THE FOCI,
WHICH ARE THE TWO FIXED POINTS INSIDE THE ELLIPSE.
NOTICE HOW BECAUSE WE HAVE A VERTICAL MAJOR AXIS,
THESE TWO POINTS ARE C UNITS ABOVE AND BELOW THE CENTER.
BUT WE'RE NOT GIVEN THE VALUE OF C, WE HAVE TO FIND IT.
WE CAN USE THE EQUATION
"A" SQUARED = B SQUARED + C SQUARED TO DETERMINE C
SINCE WE KNOW "A" SQUARED IS 25 AND B SQUARED IS 4.
LET'S GO AHEAD AND DO THIS ON THE NEXT SLIDE.
AGAIN, WE KNOW THAT "A" SQUARED MUST = B SQUARED + C SQUARED.
LET'S GO AND SOLVE THIS EQUATION FOR C SQUARED
BY SUBTRACTING B SQUARED ON BOTH SIDES.
THAT WOULD GIVE US C SQUARED = "A" SQUARED - B SQUARED
WHERE WE KNOW "A" SQUARED = 25 AND B SQUARED = 4.
SO WE WOULD HAVE C SQUARED = "A" SQUARED
OF 25 - B SQUARED OF 4.
SO WE HAVE C SQUARED = 21.
AGAIN, WE'RE ONLY CONCERNED ABOUT THE POSITIVE VALUE OF C,
SO WE'LL TAKE THE PRINCIPAL SQUARE ROOT
OF BOTH SIDES OF THE EQUATION AND WE HAVE C = SQUARE ROOT 21,
WHICH DOES NOT SIMPLIFY.
BECAUSE WE'RE GOING TO USE THIS VALUE TO PLOT ON THE POINTS
ON THE COORDINATE PLANE,
IT WOULD BE HELPFUL TO GET A DECIMAL APPROXIMATION.
SO THIS IS APPROXIMATELY 4.583.
SO NOW GOING BACK TO OUR PROBLEM, AGAIN,
WE JUST FOUND THAT C = SQUARE ROOT 21,
WHICH IS APPROXIMATELY 4.583.
AND SINCE THE TWO FOCI ARE ON THE MAJOR AXIS,
IF WE START AT THE CENTER,
WHEN WE GO UP APPROXIMATELY 4.583 UNITS OR JUST PASS 4 1/2
IT WOULD BE SOMEWHERE IN HERE.
THIS WOULD BE ONE FOCUS.
AND IF WE START AT THE CENTER
AND GO DOWN APPROXIMATELY 4.583 UNITS TO HERE,
THIS WOULD BE THE OTHER FOCUS.
NOTICE HOW TO FIND THE EXACT COORDINATES
OF THESE TWO POINTS,
IT WOULD BE DIFFICULT FROM THE COORDINATE PLANE.
SO WHAT WE'RE GOING TO DO IS ADD AND SUBTRACT SQUARE ROOT 21
FROM THE Y COORDINATE OF THE CENTER
BECAUSE THE TWO POINTS ARE ABOVE AND BELOW THE CENTER.
SO THE COORDINATES OF THE TWO FOCI WOULD BE 1
AND THEN -4 - SQUARE ROOT 21, AND THE OTHER POINT WOULD HAVE
COORDINATES (1,-4) + SQUARE ROOT 21.
AND NOW FOR THE LAST STEP,
BEFORE WE GRAPH THE ELLIPSE
WE'RE ASKED TO FIND THE ECCENTRICITY OF THE ELLIPSE,
WHICH = C DIVIDED BY "A"
WHERE C = THE SQUARE ROOT OF 21 AND "A" = 5.
SO THE SQUARE ROOT OF 21 DIVIDED BY 5.
WE'LL GET A DECIMAL APPROXIMATION,
SO APPROXIMATELY 0.917.
SO OUR ELLIPSE IS GOING TO PASS
THROUGH THE END POINTS OF THE MAJOR AND MINOR AXES
AND WOULD LOOK SOMETHING LIKE THIS.
NOTICE HOW IT IS FAIRLY ELONGATED
BECAUSE THE ECCENTRICITY IS MUCH CLOSER TO 1 THAN IT IS TO 0.
I HOPE YOU FOUND THIS HELPFUL.