## anonymous one year ago Regan is trying to find the equation of a quadratic that has a focus of (−2, 5) and a directrix of y = 13. Describe to Regan your preferred method for deriving the equation. Make sure you use Regan's situation as a model to help her understand.

1. anonymous

I don't quite understand how to find a when writing the equation.. that's all I need help with.

2. anonymous

@jchick

3. jchick

Sorry it took so long to reply lol

4. anonymous

its cool XD

5. anonymous

I waited till I saw you weren't actively helping someone

6. jchick

The definition of a parabola is that the distance of any point on it from the focus is equal to the distance of the point from the directrix. So if (x,y) is a point on the parabola, then find the distance of (x,y) from the focus (-2,5). Find the distance of (x,y) from y = 13. Equate them and you will have your quadratic equation.

7. jchick

(x+2)^2+(y−5)^2=(y−13)^2 Simplify above.

8. anonymous

but how do you find (x,y)?

9. jchick

no matter what value x is, y will always be 13

10. anonymous

why?

11. jchick

directrix y = 13 can be represented as the point (x, 13)

12. anonymous

oh okay, that makes sense.

13. jchick

So after inputting the points, you have: (x−(−2))2+(y−5)2−−−−−−−−−−−−−−−−−−√=(x−x)2+(y−13)2−−−−−−−−−−−−−−−−√

14. jchick

You can then square both sides to get: (x−(−2))2+(y−5)2= (x−x)2+(y−13)2

15. jchick

Does this make sense so far?

16. anonymous

im not sure... itdoes a little bit

17. jchick

Ok where do you need explanation?

18. jchick

You can then square both sides to get: (x−(−2))^2+(y−5)^2= (x−x)^2+(y−13)^2

19. jchick

sorry that is the correct version

20. anonymous

what are you plugging them into?

21. jchick

What do you mean?

22. jchick

Where?

23. anonymous

like are you plugging those into an equation??

24. jchick

And then that simplifies to (x+2)^2+(y−5)^2=(y−13)^2

25. jchick

Yes

26. anonymous

okay, what equation?

27. jchick

(x−(−2))2+(y−5)2−−−−−−−−−−−−−−−−−−√= (x−x)2+(y−13)2−−−−−−−−−−−−−−−−√

28. anonymous

im so confused....

29. jchick

inserted points (-2,5) and (x, 13) the distance formula: (x−x1)2+(y−y2)2−−−−−−−−−−−−−−−−−√=(x−x2)2+(y−y2)2−−−−−−−−−−−−−−−−−√

30. jchick

Do you understand now?

31. anonymous

like I was learning it like this... The graph of the quadratic is a parabola. The directrix is horizontal, so the parabola is vertical. The focus lies below the directrix, so the parabola opens downwards. General equation of a down-opening parabola:    y = a(x - h)² + k with    a < 0    vertex (h, k)    focal length p = | 1/(4a) |    focus (h, k-p)    directrix y = k+p Plug your data into the equations and solve for a now im really confused... when I was only confused about how to figure out what a is XD

32. jchick

Oh ok

33. jchick

Sorry for confusion

34. anonymous

its cool I just didn't know what you were plugging it all into.. XD

35. jchick

Having internet trouble please hang on a second

36. anonymous

okay

37. anonymous

I've never seen this done this way. Pretty cool @jchick It looks like this is using the definition of a parabola as the set of points equidistant between the focus and directrix. A circle is a set of point equidistant from the same point. The equation for a circle is $(x-h)^2+(y-k)^2=r^2$. jchick set the equations for 2 circles equal to each other, one for the focus with (-2, 5) and the other for the directrix (x, 13). That's where this came from $(x+2)^2+(y-5)^2=(x-x)^2+(y-13)$ If you mulitply everything out, and then complete the square you can find a with this method

38. anonymous

so do you use the distance formula to find a?

39. anonymous

that's pretty much what it works out to being. and you don't actually need to complete the square. a will be the coefficient of x²

40. jchick

Sorry I am back and thank you @peachpi

41. anonymous

just FOIL out the equation and you'll have an expanded equation for the parabola

42. anonymous

im still really confused... what is r in that equation??

43. anonymous

Maybe try a different method since this is confusing. We know that the vertex is halfway between the focus and directrix, so it's (-2, 9) and the parabola opens down like you said above. So the focal length is 9 - 5 = 4 |dw:1439944412409:dw|

44. anonymous

how do you find the vertex?

45. jchick

(x-a)^2 =-4p(y-b) vertex =(a,b)

46. jchick

@kaylamarie79 do you remember the standard form?

47. anonymous

The parabola opens down so the vertex has to be directly above the focus, meaning the x-coordinate has to be -2. The vertex is halfway between focus and directrix, so it's halfway between 5 and 13, so the y-coordinate has to be 9. Vertex = (-2, 9). To find a: From your list of equations above, $p=\left| \frac{ 1 }{ 4a } \right|$ Since you know p, you can solve for a.

48. anonymous

y=ax^2+bx+c

49. anonymous

@peachpi I dont know p

50. anonymous

p is the focal distance

51. anonymous

p = distance from vertex to focus = distance from vertex to directrix = half the distance from focus to directrix

52. anonymous

1 min im trying wrap my mind around this..

53. anonymous

what are you thinking?

54. anonymous

all I can think is i'm sooo confused... like I don't understand where your getting a the focal point(p) or the vertex... and I keep getting more confused as y'all repeat things.. like i'm terrible in math and have to see every single step or the entire equation gets jumbled around in my head...

55. anonymous

ok. starting from the beginning. The directrix is y = 13. The focus is (-2, 5).|dw:1439945735884:dw|

56. anonymous

uh- huh

57. anonymous

The vertex has to be halfway between them. 9 is halfway between 5 and 13, so the vertex has to be (-2, 9)|dw:1439945891758:dw|

58. anonymous

how do you get the vertex point?

59. anonymous

what number is halfway between 5 and 13?

60. anonymous

oh okay. how do you know the -2 though?

61. anonymous

because the parabola opens down, the focus and vertex must have the same x-coordinate

62. anonymous

is it always equal as the focus or only when opening down?

63. anonymous

only when opening up or down. When it opens left or right the y-coordinates are the same

64. anonymous

oh okay that makes sense

65. anonymous

that's the vertex. Now for p. That's the distance between the focus and vertex, so it's p = 9 - 5.

66. anonymous

p = 4

67. anonymous

with me?

68. anonymous

yeah, so to get a you use the focal length equation but just plug in p?

69. anonymous

yes. but since it opens down make p be -4 and you can drop the absolute value signs

70. anonymous

ohhhhh see the other ways just got really confusing..

71. anonymous

okk cool

72. jchick

73. anonymous

yeah, I did.., I understand now XD

74. jchick

Ok good

75. anonymous

thank you guys for helping

76. jchick

Yeah no problem and sorry about being confusing lol