- GIVEN THE EQUATION OF THE PARABOLA,
WE WANT TO FIND ALL OF THE FOLLOWING CHARACTERISTICS
BY WRITING THE EQUATION IN STANDARD FORM
OR THIS FORM HERE.
NOTICE HOW THE LEFT SIDE OF THIS EQUATION
WOULD BE A PERFECT SQUARE TRINOMIAL IN TERMS OF X,
AND THE RIGHT SIDE IN TERMS OF Y IS FACTORED.
SO FOR THE FIRST STEP,
WE'LL HAVE TO COMPLETE THIS SQUARE ON X SQUARED + 6
TO FORM A PERFECT SQUARE TRINOMIAL.
SO WE WOULD HAVE X SQUARED + 6X + A CONSTANT
= 4Y + 11 + A CONSTANT.
SO TO COMPLETE THIS SQUARE,
WE'RE GOING TO TAKE HALF OF THE COEFFICIENT OF X,
OR HALF OF 6, AND THEN SQUARE IT.
WELL, HALF OF 6 IS 3, 3 SQUARED IS EQUAL TO 9,
SO WE'RE GOING TO ADD NINE TO BOTH SIDES OF THE EQUATION.
NOTICE HOW NOW WE CAN FACTOR THE LEFT SIDE.
WE'D HAVE X AND X AND THE FACTORS OF 9 THAT ADD TO 6
ARE +3 AND +3,
SO WE HAVE + 3 HERE AND + 3 HERE.
AND ON THE RIGHT SIDE WE'D HAVE 4Y + 20,
SO NOW ON THE LEFT WE CAN WRITE
THE QUANTITY X + 3 SQUARED EQUALS
WE'RE GOING TO FACTOR OUT 4,
SO WE'LL HAVE 4 x THE QUANTITY Y + 5.
NOW THAT WE HAVE THE EQUATION IN THE CORRECT FORM,
THE FIRST THING WE SHOULD RECOGNIZE
IS THAT THE X PART OF THE EQUATION IS SQUARED
AND THEREFORE WE'LL HAVE A VERTICAL AXIS,
OR VERTICAL AXIS OF SYMMETRY,
WHICH MEANS THE PARABOLA WILL OPEN UP OR DOWN
BASED UPON THE VALUE OF "A".
IF "A" IS POSITIVE THE PARABOLA OPENS UP,
IF "A" IS NEGATIVE THE PARABOLA OPENS DOWN.
FOR THIS EXAMPLE, NOTICE THAT 4A = +4,
THEREFORE "A" = 1, OR +1,
THEREFORE WE KNOW THE PARABOLA OPENS UP.
SO GOING BACK TO OUR EQUATION NOW,
LETS FIRST FIND THE VERTEX.
THE COORDINATES OF THE VERTEX WILL BE (H, K).
SO IF WE HAVE THE QUANTITY X + 3,
THE X COORDINATE OF THE VERTEX WOULD BE -3,
AND IF WE HAVE THE QUANTITY Y + 5,
THE Y COORDINATE WOULD BE -5.
SO LET'S GO AHEAD AND PLOT THE VERTEX
ON THE COORDINATE PLANE (-3, -5) WOULD BE HERE.
LET'S PUT A V FOR VERTEX.
NOW, WE KNOW THIS PARABOLA OPENS UP,
SO WE'RE GOING TO MAKE A ROUGH SKETCH
OF A PARABOLA OPENING UP
BECAUSE IT WILL HELP US FIND ALL THIS OTHER INFORMATION.
SO IT WOULD LOOK SOMETHING LIKE THIS,
BUT BECAUSE WE KNOW IT OPENS UP,
WE NOW KNOW THAT THE FOCUS WOULD BE THE VERTEX,
OR INSIDE THE PARABOLA
AND WE KNOW THE DIRECTRIX WOULD BE BELOW THE VERTEX.
AND NOW, THE MOST IMPORTANT THING TO REMEMBER
IS THE DISTANCE FROM THE VERTEX TO THE FOCUS
IS EQUAL TO THE ABSOLUTE VALUE OF "A",
AND THE DISTANCE FROM THE VERTEX TO THE DIRECTRIX
IS ALSO EQUAL TO THE ABSOLUTE VALUE OF "A".
AND SINCE "A" = 1,
OF COURSE THE ABSOLUTE VALUE OF ONE IS STILL ONE,
SO NOW IF WE GO BACK TO THE GRAPH--THIS IS OUR VERTEX
AND THE FOCUS IS ABOVE THE VERTEX,
IT MUST BE ONE UNIT ABOVE OR THIS POINT HERE,
WHICH MEANS TO FIND THE COORDINATES OF THE FOCUS
WE'LL ADD ONE TO THE Y COORDINATE OF THE VERTEX
BECAUSE IT'S ONE UNIT ABOVE THE VERTEX,
WHICH MEANS THE FOCUS WILL HAVE COORDINATES -3,
THEN -5 + 1 WOULD BE -4, WHICH WE CAN VERIFY FROM THE GRAPH.
AND THEN THE DIRECTRIX WOULD BE ONE UNIT
BELOW THE VERTEX OR A HORIZONTAL LINE
PASSING THROUGH THIS POINT HERE.
SO THIS WOULD BE OUR DIRECTRIX,
AND BECAUSE THIS IS A HORIZONTAL LINE
WE KNOW THE EQUATION OF THIS LINE WOULD BE Y = -6.
OF COURSE, ANOTHER WAY TO OBTAIN THIS EQUATION
WOULD BE TO SUBTRACT ONE
FROM THE Y COORDINATE OF THE VERTEX.
NOTICE THAT -5 - 1 IS -6,
WHICH WOULD GIVE US OUR EQUATION.
AND THEN FOR THE EQUATION OF THE AXIS OF SYMMETRY, OR AXIS,
IT WOULD BE A VERTICAL LINE PASSING THROUGH THE FOCUS
AND THE VERTEX OR THIS VERTICAL LINE HERE.
AND AGAIN, BY GRAPHING THIS
WE CAN TELL THE EQUATION OF THE AXIS WOULD BE X = -3.
NOTICE, THE FORMULA SAYS IT'S X = H,
WHERE H IS THE X COORDINATE OF THE VERTEX,
NOTICE HOW THE VERTEX DOES HAVE AN X COORDINATE OF -3.
AND THEN FINALLY,
WE'LL FIND THE ENDPOINTS OF THE LATUS RECTUM,
WHICH WILL GIVE US 2 MORE POINTS ON THE PARABOLA
SO WE CAN THEN SKETCH A MORE ACCURATE GRAPH
OF THE PARABOLA.
SO THE LATUS RECTUM IS THIS SEGMENT HERE,
WHERE THE MIDPOINT IS THE FOCUS,
THE 2 ENDPOINTS ARE ON THE PARABOLA,
AND THE LATUS RECTUM IS PARALLEL TO THE DIRECTRIX.
MORE IMPORTANTLY, IT HAS A LENGTH
= TO THE ABSOLUTE VALUE OF 4A.
SO LOOKING AT OUR EQUATION NOTICE THAT 4A IS = TO 4,
AND THEREFORE THE LATUS RECTUM HAS A LENGTH OF FOUR UNITS.
LET'S GO AHEAD AND CLEAR THIS ROUGH SKETCH OF OUR PARABOLA,
BECAUSE THE FOCUS IS THE MIDPOINT OF THE LATUS RECTUM,
WE KNOW THAT IF THE LENGTH IS FOUR UNITS,
ONE ENDPOINT WOULD BE TWO UNITS
TO THE RIGHT OF THE FOCUS, OR THIS POINT HERE,
AND THE OTHER ENDPOINT WOULD BE TWO UNITS
TO THE LEFT OF THE FOCUS OR THIS POINT HERE.
NOTICE HOW THIS SEGMENT WOULD HAVE A LENGTH OF FOUR UNITS,
IT PASSES THROUGH THE FOCUS
AND IT'S PARALLEL TO THE DIRECTRIX.
SO TO FIND THE COORDINATES WE COULD JUST LOOK AT THE GRAPH.
NOTICE HOW THE LEFT POINT WOULD BE (-5, -4)
AND THE SECOND ENDPOINT WOULD BE (-1, -4).
OF COURSE, THE OTHER OPTION IS TO ADD AND SUBTRACT TWO
FROM THE X COORDINATE OF THE FOCUS.
NOTICE THAT -3 - 2 WOULD BE -5, AND -3 + 2 = -1.
BUT MORE IMPORTANTLY,
NOW THAT WE HAVE THREE POINTS ON THE PARABOLA
WE CAN SKETCH A MORE ACCURATE PARABOLA
PASSING THROUGH THE VERTEX
AND THE TWO ENDPOINTS OF THE LATUS RECTUM,
OR FOCAL DIAMETER.
SO THE PARABOLA WOULD LOOK SOMETHING LIKE THIS.
OF COURSE, WE COULD MAKE A TABLE OF VALUES,
OR FIND THE INTERCEPTS,
IF WE WANTED MORE POINTS ON THE PARABOLA.
I HOPE YOU FOUND THIS EXPLANATION HELPFUL.