anonymous
  • anonymous
Medal & fan Using a directrix of y = -2 and a focus of (1, 6), what quadratic function is created?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Nnesha @zepdrix @MeganEdward
anonymous
  • anonymous
or @dan815 or @merchandize lol
Nnesha
  • Nnesha
|dw:1444448203481:dw| ;-;

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Jhannybean
  • Jhannybean
Recall that the vertex of a parabola lies halfway between the focus and the directrix
Nnesha
  • 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
  • Nnesha
npI y-c I is same as (y-c)^2 and (a,b) is the focus point any question ?
anonymous
  • 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
  • Nnesha
i see sorry directrix is a line y = c |dw:1444448897371:dw| and the focus point is at (1,6)
dan815
  • dan815
http://prntscr.com/8ppjc9
Nnesha
  • Nnesha
ahh distance between the point `on parabola ` i missd that part :(
dan815
  • dan815
http://prntscr.com/8ppjog
anonymous
  • anonymous
I'm sorry, I just dont get it :(
anonymous
  • anonymous
Ok mabey I can help
Nnesha
  • 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
  • anonymous
Right
Nnesha
  • Nnesha
now draw focus point (1,6) |dw:1444449417779:dw| read the definition on the dan's secreenshot :P |dw:1444449585947:dw|
anonymous
  • anonymous
Okay, I get everything up to this point
Nnesha
  • 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
  • anonymous
Okay is there a formula for this?
Nnesha
  • 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
  • Nnesha
(a,b) is the focus point c= directrix
Nnesha
  • Nnesha
now plugin the numbers (a,b) = (1,6) c= -2
anonymous
  • anonymous
so sqrt of (x - 1)^2 + (y - 6)^2 = (y + 2)^2 right?
Nnesha
  • 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
  • anonymous
Ill let you know, thanks for helping! And what do you mean by "take square both sides" ?
Nnesha
  • Nnesha
when we take square both sides tit would be \[\huge\rm (x-1)^2 +(x-6)^2 = (y+2)^2\]
anonymous
  • anonymous
ohh okay, right. to get rid of the sqrt and the absolutes
Nnesha
  • Nnesha
yes right
anonymous
  • anonymous
and then what??
Nnesha
  • 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
  • Nnesha
ugh typo it's \[\huge\rm (x-1)^2 +(y-6)^2 = (y+2)^2\]
anonymous
  • anonymous
do i foil x-1 ^2 too?
Nnesha
  • Nnesha
sorry sorry :( (y-6)(y-6)
Nnesha
  • 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
  • anonymous
okay so (y^2 - 12y + 36) + (x-1)^2 = y^2 +4y + 4
Nnesha
  • Nnesha
right now combine like terms |dw:1444450916236:dw|
Nnesha
  • Nnesha
|dw:1444450996389:dw|
Nnesha
  • Nnesha
sorry about the cool writing ,-,
anonymous
  • anonymous
wait, that last thing threw me off
Nnesha
  • Nnesha
ohh okay which one please ask :=)
Nnesha
  • Nnesha
i just moved all terms to the right side except (x-1)^2
anonymous
  • anonymous
oh never mind, i get it! Sorry
Nnesha
  • Nnesha
okay thats fine
anonymous
  • anonymous
Okay and now what do i do?
Nnesha
  • Nnesha
let me think ... :( i'm lost :( wait a sec plz
anonymous
  • anonymous
Okkay thats fine!
anonymous
  • 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
  • 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
  • Nnesha
that's 16 i'm so sorry
Nnesha
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.