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- WE WANT TO FIND THE STANDARD EQUATION OF THE ELLIPSE
WITH FOCI (4,0) AND (-4,0),
AND THE LENGTH OF THE MINOR AXIS IS 6 UNITS.
THE STANDARD EQUATION OF AN ELLIPSE
WILL TAKE ON TWO FORMS
BASED UPON WHETHER WE HAVE A VERTICAL MAJOR AXIS
OR A HORIZONTAL MAJOR AXIS.
IN BOTH CASES A IS GREATER THAN B,
AND THEREFORE A SQUARED IS GREATER THAN B SQUARED,
MEANING A SQUARED IS THE LARGER DENOMINATOR.
SO IF THE LARGER DENOMINATOR IS UNDER THE Y PART
OF THE EQUATION, AS WE SEE HERE,
WE'LL HAVE A VERTICAL MAJOR AXIS.
AND IF THE LARGER DENOMINATOR
IS UNDER THE X PART OF THE EQUATION, AS WE SEE HERE,
WE'LL HAVE A HORIZONTAL MAJOR AXIS.
REMEMBER THE MAJOR AXIS IS THE LONGER AXIS,
WHICH THIS ELLIPSE WOULD BE THIS AXIS HERE,
AND THE MINOR AXIS IS THE SHORTER AXIS,
WHICH SHOULD BE THIS AXIS HERE.
NOTICE THE DISTANCE FROM CENTER
TO THE TWO END POINTS OF THE MAJOR AXIS,
ALSO CALLED 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
WHERE THE CENTER IS THE MID POINT OF BOTH AXIS.
SO LET'S GO AHEAD AND START BY GRAPHING
THE GIVEN INFORMATION ON THE COORDINATE PLANE
AND SEE WHAT INFORMATION WE CAN FIND.
THE TWO FOCI ARE (4,0), WHICH WOULD BE HERE,
AND (-4,0) WHICH WOULD BE HERE.
WELL, WE SHOULD RECOGNIZE THAT THE FOCI ARE ON THE MAJOR AXIS,
AND THEREFORE WE'RE GOING TO HAVE A HORIZONTAL MAJOR AXIS.
SO WE'LL BE USING THIS EQUATION HERE FOR OUR ELLIPSE.
NOTICE THE CENTER IS ALSO HALFWAY BETWEEN THE TWO FOCI,
SO THIS POINT HERE WOULD BE OUR CENTER.
SO FOR THIS EXAMPLE THE CENTER IS AT THE ORIGIN.
AND THEN FINALLY WE'RE TOLD THE LENGTH OF THE MAJOR AXIS
IS SIX UNITS,
WHICH MEANS 2B IS = TO 6 OR B IS = TO 3.
SO TO FIND THE ENDPOINTS OF THE MINOR AXIS
STARTING AT THE CENTER, WE WOULD GO UP 3 UNITS.
THIS WOULD BE ONE ENDPOINT,
AND DOWN 3 UNITS FROM THE CENTER FOR THE OTHER ENDPOINT.
SO THIS AXIS HERE WOULD BE THE MINOR AXIS.
SO NOW LET'S GO AND RECORD ALL THE INFORMATION
WE CAN FROM OUR SKETCH.
WELL, FIRST WE KNOW THAT THE CENTER IS AT THE ORIGIN,
SO THE COORDINATES WOULD BE (0,0).
AND LOOKING AT OUR EQUATION,
THE CENTER HAS COORDINATES (H,K).
SO H IS = TO 0 AND K IS = TO 0.
NEXT, WE ALREADY SAID THAT THE DISTANCE FROM THE CENTER
TO THE ENDPOINTS OF THE MINOR AXIS IS EQUAL TO B UNITS,
AND SINCE THIS DISTANCE IS = TO 3 WE KNOW B IS = TO 3.
AND ALSO THE DISTANCE FROM THE CENTER TO THE TWO FOCI
IS EQUAL TO C UNITS.
AND SINCE THIS DISTANCE HERE IS = TO 4 UNITS
WE KNOW C IS = TO 4.
SO TO WRITE THE EQUATION OF OUR ELLIPSE
WE ALSO HAVE TO FIND A SQUARED,
WHICH WE CAN DO USING THE EQUATION
A SQUARED = B SQUARED + C SQUARED,
WHICH IS ALWAYS TRUE FOR AN ELLIPSE.
SO WE'RE GOING TO HAVE A SQUARED IS = TO B SQUARED.
AND SINCE B IS 3, 3 SQUARED IS 9 + C SQUARED.
WELL, C IS 4 AND 4 SQUARED IS 16.
SO WE HAVE A SQUARED = 9 + 16 OR 25,
WHICH IS ALL WE NEED TO WRITE THE EQUATION OF OUR ELLIPSE.
WE KNOW A SQUARED IS 25.
IF B IS = TO 3 THEN B SQUARED IS = TO 9.
SO OUR EQUATION OF OUR ELLIPSE WILL BE IN THIS FORM HERE WHERE,
AGAIN H AND K ARE 0.
SO WE WOULD HAVE X SQUARED DIVIDED BY A SQUARED,
WHICH IS 25, + Y SQUARED DIVIDED BY B SQUARED,
WHICH IS 9, MUST = 1.
AGAIN, NOTICE HOW THE LARGER DENOMINATOR
OR A SQUARED IS UNDER THE X PART OF THE EQUATION
BECAUSE WE KNOW WE HAVE A HORIZONTAL MAJOR AXIS.
SO THIS WOULD BE THE EQUATION OF OUR ELLIPSE.
LET'S GO AHEAD AND FINISH BY GRAPHING OUR ELLIPSE.
NOTICE THAT WE'D ALSO STILL WANT TO FIND THE TWO ENDPOINTS
OF THE MAJOR AXIS
SO WE CAN MAKE AN ACCURATE SKETCH.
AND THE TWO ENDPOINTS OF THE MAJOR AXIS, OR THE VERTICES,
WILL BE A UNITS TO THE RIGHT AND LEFT OF THE CENTER
SINCE WE HAVE A HORIZONTAL MAJOR AXIS.
AND SINCE A SQUARED IS = TO 25 AND WE'RE ONLY CONCERNED
ABOUT THE POSITIVE VALUE OF A, WE KNOW THAT A MUST = +5.
WHICH MEANS THE TWO ENDPOINTS OF THE MAJOR AXIS
WILL BE FIVE UNITS TO THE RIGHT AND LEFT OF THE CENTER.
SO ONE ENDPOINT WOULD BE HERE,
THE OTHER ENDPOINT WOULD BE HERE.
SO OUR ELLIPSE WOULD LOOK SOMETHING LIKE THIS.
OKAY, I HOPE YOU FOUND THIS EXPLANATION HELPFUL.