Medal & fan Using a directrix of y = -2 and a focus of (1, 6), what quadratic function is created?

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Medal & fan Using a directrix of y = -2 and a focus of (1, 6), what quadratic function is created?

Mathematics
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|dw:1444448203481:dw| ;-;

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Recall that the vertex of a parabola lies halfway between the focus and the directrix
alright let's pick point (x,y) |dw:1444448374229:dw| we need to find distance between the point (x,y) and focus and the distance between point and directirx \[\sqrt{(x - a)^2+(y -b)^2}= \left| y-c \right|\] \[\sqrt{(x- 1)^2 + (y-6)^2} = (y-(-2))^2\]
npI y-c I is same as (y-c)^2 and (a,b) is the focus point any question ?
Yes, I'm just completely lost :/ Don't understand this question as well as the other one.. I dont remember what a directrix even is
i see sorry directrix is a line y = c |dw:1444448897371:dw| and the focus point is at (1,6)
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ahh distance between the point `on parabola ` i missd that part :(
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I'm sorry, I just dont get it :(
Ok mabey I can help
y = c ( it's horizontal line ) so y =-2 draw a horizontal line at -2 |dw:1444449356899:dw| red line is directrix right
Right
now draw focus point (1,6) |dw:1444449417779:dw| read the definition on the dan's secreenshot :P |dw:1444449585947:dw|
Okay, I get everything up to this point
to write an equation we have to find 1) distance between the point `on parabola ` to the focus 2nd) distance between the point " on parabola" to the directrix
Okay is there a formula for this?
|dw:1444449758151:dw| blue line distance ( between point and directrix ) red one distance between (point and focus ) and the equation i posted above \[\huge\rm \sqrt{(x-a)^2 +(y-b)^2 }= \left| y-c \right|\]
(a,b) is the focus point c= directrix
now plugin the numbers (a,b) = (1,6) c= -2
so sqrt of (x - 1)^2 + (y - 6)^2 = (y + 2)^2 right?
hmm \[\huge\rm \sqrt{(x-1)^2 +(y-6)^2 }= \left| y+2 \right|\] take square both sides please let me know if you don't get anything i'm sleepy e,e
Ill let you know, thanks for helping! And what do you mean by "take square both sides" ?
when we take square both sides tit would be \[\huge\rm (x-1)^2 +(x-6)^2 = (y+2)^2\]
ohh okay, right. to get rid of the sqrt and the absolutes
yes right
and then what??
and then foil (x-6)^2 is same as (x-6)(x-6) same with (y+2)^2 bec as dan mentioned we need an equation y= form
ugh typo it's \[\huge\rm (x-1)^2 +(y-6)^2 = (y+2)^2\]
do i foil x-1 ^2 too?
sorry sorry :( (y-6)(y-6)
well i'll saw leave it as (x-h)^2 we can write an equation in vertex form \[y=a(x-h)^2+k\]
okay so (y^2 - 12y + 36) + (x-1)^2 = y^2 +4y + 4
right now combine like terms |dw:1444450916236:dw|
|dw:1444450996389:dw|
sorry about the cool writing ,-,
wait, that last thing threw me off
ohh okay which one please ask :=)
i just moved all terms to the right side except (x-1)^2
oh never mind, i get it! Sorry
okay thats fine
Okay and now what do i do?
let me think ... :( i'm lost :( wait a sec plz
Okkay thats fine!
I just submitted the exam because i have to go to bed. I got the answer wrong too :/ but thanks for all your help!
alright i'll going to expand (x-1)^2 \[\large\rm x^2-2x+1 +y^2-12y+36=y^2+4y+4\] |dw:1444451678408:dw|
that's 16 i'm so sorry
i shouldn't start helping u so my fault i apologize i'll look at it tomorrow will try to explain and i wil lfocus on this

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