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

- anonymous

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

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- schrodinger

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- anonymous

@Nnesha @zepdrix @MeganEdward

- anonymous

or @dan815 or @merchandize lol

- Nnesha

|dw:1444448203481:dw| ;-;

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## More answers

- Jhannybean

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

- Nnesha

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
\[\sqrt{(x - a)^2+(y -b)^2}= \left| y-c \right|\] \[\sqrt{(x- 1)^2 + (y-6)^2} = (y-(-2))^2\]

- Nnesha

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

- anonymous

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

- Nnesha

i see sorry
directrix is a line y = c
|dw:1444448897371:dw|
and the focus point is at (1,6)

- dan815

http://prntscr.com/8ppjc9

- Nnesha

ahh distance between the point `on parabola ` i missd that part :(

- dan815

http://prntscr.com/8ppjog

- anonymous

I'm sorry, I just dont get it :(

- anonymous

Ok mabey I can help

- Nnesha

y = c ( it's horizontal line )
so y =-2 draw a horizontal line at -2 |dw:1444449356899:dw|
red line is directrix right

- anonymous

Right

- Nnesha

now draw focus point (1,6) |dw:1444449417779:dw|
read the definition on the dan's secreenshot :P |dw:1444449585947:dw|

- anonymous

Okay, I get everything up to this point

- Nnesha

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

- anonymous

Okay is there a formula for this?

- Nnesha

|dw:1444449758151:dw|
blue line distance ( between point and directrix )
red one distance between (point and focus )
and the equation i posted above \[\huge\rm \sqrt{(x-a)^2 +(y-b)^2 }= \left| y-c \right|\]

- Nnesha

(a,b) is the focus point
c= directrix

- Nnesha

now plugin the numbers
(a,b) = (1,6)
c= -2

- anonymous

so sqrt of (x - 1)^2 + (y - 6)^2 = (y + 2)^2 right?

- Nnesha

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

- anonymous

Ill let you know, thanks for helping! And what do you mean by "take square both sides" ?

- Nnesha

when we take square both sides tit would be \[\huge\rm (x-1)^2 +(x-6)^2 = (y+2)^2\]

- anonymous

ohh okay, right. to get rid of the sqrt and the absolutes

- Nnesha

yes right

- anonymous

and then what??

- Nnesha

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

- Nnesha

ugh typo it's \[\huge\rm (x-1)^2 +(y-6)^2 = (y+2)^2\]

- anonymous

do i foil x-1 ^2 too?

- Nnesha

sorry sorry :( (y-6)(y-6)

- Nnesha

well i'll saw leave it as (x-h)^2 we can write an equation in vertex form \[y=a(x-h)^2+k\]

- anonymous

okay so
(y^2 - 12y + 36) + (x-1)^2 = y^2 +4y + 4

- Nnesha

right now combine like terms |dw:1444450916236:dw|

- Nnesha

|dw:1444450996389:dw|

- Nnesha

sorry about the cool writing ,-,

- anonymous

wait, that last thing threw me off

- Nnesha

ohh okay which one please ask :=)

- Nnesha

i just moved all terms to the right side except (x-1)^2

- anonymous

oh never mind, i get it! Sorry

- Nnesha

okay thats fine

- anonymous

Okay and now what do i do?

- Nnesha

let me think ... :( i'm lost :( wait a sec plz

- anonymous

Okkay thats fine!

- anonymous

I just submitted the exam because i have to go to bed. I got the answer wrong too :/ but thanks for all your help!

- Nnesha

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|

- Nnesha

that's 16
i'm so sorry

- Nnesha

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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