## anonymous one year ago Find the standard form of the equation of the parabola with a focus at (-4, 0) and a directrix at x = 4. Could someone please help?

1. wintersuntime

Write an equation for the parabola with focus (4,0) and directrix y=2. Thanks! .. Standard form of parabola: (x-h)^2=4p(y-k), with (h,k) being the (x,y) coordinates of the vertex Given parabola opens downward with axis of symmetry at x=4. Vertex at (4,1). p=1 Equation of parabola: (x-4)^2=-4(y-1) (ans) see the graph below as a visual check on the answer .. y=1-(x-4)^2/4

2. anonymous

I'm so confused I'm sorry @wintersuntime

3. wintersuntime

Are you trying to graph it ?

4. wintersuntime

And it's okay I love helping others

5. anonymous

No. These are the answer I am given: A. y^2 = -8x B. 16y = x^2 C. y = -1/16x^2 D. x = -1/16y^2

6. wintersuntime

What do you think the answer will be ?

7. anonymous

I am thinking C. @wintersuntime

8. wintersuntime

why ?

9. anonymous

Honestly, I'm not sure. I just thought that standard form looked most similar to that choice. @wintersuntime

10. anonymous

its not C

11. anonymous

Could you explain how to find the correct answer then? I am so lost. @saseal

12. anonymous

|dw:1438996328137:dw|

13. anonymous

the parabola is opening down because the focus is below the directrix

14. anonymous

But the parabola's focus is at -4, 0. So would that mean the parabola's focus would be shifted to the left 4? Not down 4?

15. anonymous

no, focus has nothing to do with shifting

16. anonymous

Yes, but you put the point at (0, -4) it looks like.

17. anonymous

lel i made a mistake at the drawing suppose to be (0,p)

18. anonymous

Ok, but I'm still confused as to what the standard form would look like?

19. anonymous

the standard form flavor looks like this $x^2=4py$ or $y^2=4px$

20. anonymous

depends on where your parabola opens out

21. anonymous

Ok so since the parabola opens down, would it be y^2 = 4px?

22. anonymous

|dw:1438996750393:dw|

23. anonymous

|dw:1438996883193:dw|

24. anonymous

so its not y^2

25. anonymous

Oh ok, so it would be in the form of x^2 = 4py?

26. anonymous

yes

27. anonymous

So the answer then would be B? @saseal

28. Loser66

Tip: no need to consider whether p >0 or p <0, if you see the focus, just plug that p in For example, on your problem, focus is (0,-4) , hence plug -4 in the form x^2 = 4 p y x^2 = 4*(-4) y.

29. Loser66

then isolate y to get the form.

30. anonymous

should be x^2 = -16y

31. anonymous

That's not the answer. I just checked.

32. anonymous

I think it's y = -1/16x^2

33. Loser66

surely not $$x^2 = -16y$$ divided both sides by -16, you get y = -(1/16)x^2

34. anonymous

Ok I was correct. Thank you!

35. anonymous

I wish I could give you all medals.

36. Loser66

give it to saseal, I have more than 3 thousands of it. hehehe.

37. anonymous

lol

38. anonymous

Could you guys just check a couple for me? I'm really confused on these, and have them done, just want to make sure they're correct.

39. anonymous

alright

40. anonymous

Find the vertex, focus, directrix, and focal width of the parabola. x = 3^y2 Would the answer be: Vertex: (0,0) Focus: (1/12,0) Focus Width: .333 or 1/12 Directrix: -1/12

41. anonymous

$x=3y^2$$y^2=\frac{ 1 }{ 3 }x$$y^2 = 4(\frac{ 1 }{ 12 }) x$

42. anonymous

Ok...? So I'm incorrect?

43. anonymous

focus width is wrong and directrix usually they write it as directrix = x = -1/12 so people know which axis it is on

44. anonymous

Oh ok sorry. But my choice for the vertex, focus, and directrix to all remain the same, the focal width has to be .33

45. anonymous

yes

46. anonymous

focal width aka latus rectum = 4p

47. anonymous

That's not one of my choices. I think they wanted you to solve for the width.

48. anonymous

|dw:1438997923182:dw|

49. anonymous

Ok thanks!

50. anonymous

For: A building has an entry the shape of a parabolic arch 74 ft high and 28 ft wide at the base as shown below. Would my answer be: y = −37/98x^2

51. anonymous

|dw:1438998052704:dw|

52. anonymous

Ok

53. anonymous

so its definitely p<0

54. anonymous

So what would that mean for the ending answer?

55. Loser66

To this kind of problem, you need derive from the standard form Vertex (0,74), Two points are (-14,0) and (14,0) ok?

56. anonymous

Ok

57. Loser66

Vertex form is $$y = a(x-h)^2 +k$$ $$y = ax^2 + 74$$

58. anonymous

|dw:1438998433802:dw|

59. anonymous

always better t o draw it out haha

60. Loser66

Now, pick one of the point, I pick (14,0) , that is x =14 , y =0 and plug them in $$0= a (14)^2 +74$$ $$a(14)^2 = -74$$ $$a = \dfrac{-74}{196}$$

61. Loser66

Now, replace back $$y = \dfrac{-74}{196}x^2$$ @saseal they didn't say the base of the parabola is the line passed through focus. You cannot apply the form of focus, directrix here.

62. Loser66

@jdoherty Got what I meant?

63. anonymous

Ok, so my answer would be correct then. I do understand. (:

64. anonymous

yea ikr thats why i didnt use anything to do with focus

65. Loser66

66. anonymous

But I just simplified the fraction, dividing both the top and bottom by 2... And I just have one more question. Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7). Those still are the most confusing to me.

67. anonymous

x^2 = 4 (-7)y

68. Loser66

When they give you the focus or directrix, they want you to define the form of the parabola. You know that the focus is ALWAYS inside of the parabola, right? hence ,your parabola is up or down?

69. anonymous

So it would be: y = -1/28 x^2?

70. anonymous

@jdoherty refer to the drawings above to see where the parabola opens up

71. anonymous

I did refer to that, but based on the equation, I got my answer of y = -1/28x^2

72. anonymous

@saseal

73. anonymous

yea?

74. anonymous

so thats correct

75. anonymous

sorry i fell asleep