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

1. anonymous

@Nnesha @zepdrix @MeganEdward

2. anonymous

or @dan815 or @merchandize lol

3. Nnesha

|dw:1444448203481:dw| ;-;

4. Jhannybean

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

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

6. Nnesha

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

7. 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

8. Nnesha

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

9. dan815
10. Nnesha

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

11. dan815
12. anonymous

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

13. anonymous

Ok mabey I can help

14. Nnesha

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

15. anonymous

Right

16. Nnesha

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

17. anonymous

Okay, I get everything up to this point

18. 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

19. anonymous

Okay is there a formula for this?

20. Nnesha

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

21. Nnesha

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

22. Nnesha

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

23. anonymous

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

24. 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

25. anonymous

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

26. Nnesha

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

27. anonymous

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

28. Nnesha

yes right

29. anonymous

and then what??

30. 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

31. Nnesha

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

32. anonymous

do i foil x-1 ^2 too?

33. Nnesha

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

34. Nnesha

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

35. anonymous

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

36. Nnesha

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

37. Nnesha

|dw:1444450996389:dw|

38. Nnesha

sorry about the cool writing ,-,

39. anonymous

wait, that last thing threw me off

40. Nnesha

41. Nnesha

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

42. anonymous

oh never mind, i get it! Sorry

43. Nnesha

okay thats fine

44. anonymous

Okay and now what do i do?

45. Nnesha

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

46. anonymous

Okkay thats fine!

47. 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!

48. 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|

49. Nnesha

that's 16 i'm so sorry

50. 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