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or @dan815 or @merchandize lol

|dw:1444448203481:dw| ;-;

Recall that the vertex of a parabola lies halfway between the focus and the directrix

npI y-c I is same as (y-c)^2 and (a,b) is the focus point
any question ?

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

Right

Okay, I get everything up to this point

Okay is there a formula for this?

(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?

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??

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!

that's 16
i'm so sorry