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Mimi_x3

  • 3 years ago

If the chord of contact of the tangents to a parabola x^2=4ay, from an external point T(x1,y1) meets the directrix at R, prove that RT subtends a right angle at the focus.

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  1. perl
    • 3 years ago
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    mmmmm, you werent kidding around

  2. Mimi_x3
    • 3 years ago
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    Is this easy for you ? xD

  3. perl
    • 3 years ago
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    i will have to think about it

  4. perl
    • 3 years ago
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    the directrix is that line under the parabola

  5. Mimi_x3
    • 3 years ago
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    lols, you were thinking that its easy ? xD

  6. Mimi_x3
    • 3 years ago
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    Yup, y=-a

  7. perl
    • 3 years ago
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    what class is this ?

  8. Mimi_x3
    • 3 years ago
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    I live in Australia, so its kinda different, there's only a maths class xD

  9. perl
    • 3 years ago
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    sorry wrong question, have you done calculus

  10. Mimi_x3
    • 3 years ago
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    Yes I have, how did you know ? xD

  11. Mimi_x3
    • 3 years ago
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    What grade are you in ?

  12. perl
    • 3 years ago
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    it will definitely help

  13. perl
    • 3 years ago
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    im a sophomore in college

  14. Mimi_x3
    • 3 years ago
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    Cool~ college, and you havent learnt that ?

  15. perl
    • 3 years ago
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    not specifically, but i bet i can do this

  16. Mimi_x3
    • 3 years ago
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    lols, then try hehe

  17. perl
    • 3 years ago
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    im trying to make a diagram, im not even sure i understand the question yet

  18. perl
    • 3 years ago
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    ok you have an external pt, and you want the tangent through it

  19. Mimi_x3
    • 3 years ago
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    Yeah, I don't really know where's the external point is, thats why im lost =/

  20. perl
    • 3 years ago
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    ok first thing is first. the chord of contact means something special

  21. Mimi_x3
    • 3 years ago
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    Something special ? What do you mean ?

  22. perl
    • 3 years ago
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    there is a picture of a chord of contacts http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf

  23. perl
    • 3 years ago
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    well given an external point there are two tangents from that point to the parabola

  24. perl
    • 3 years ago
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    now where the tangents intersect the parabola, now draw a chord through those two points

  25. perl
    • 3 years ago
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    where the tangents meet the parabola, the two tangents

  26. Mimi_x3
    • 3 years ago
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    wow~ nice pdf file, its detailed on parametrics ty

  27. Mimi_x3
    • 3 years ago
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    But where is the external point ?

  28. perl
    • 3 years ago
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    anywhere

  29. Mimi_x3
    • 3 years ago
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    Can't it have to be specific

  30. perl
    • 3 years ago
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    in the pdf, the point is below the parabola

  31. Mimi_x3
    • 3 years ago
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    It can be but it doesn't make sense when it says RT subtends a right angle at the focus.

  32. perl
    • 3 years ago
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    like this

  33. Mimi_x3
    • 3 years ago
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    like this ?

  34. perl
    • 3 years ago
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    |dw:1319452686670:dw|

  35. Mimi_x3
    • 3 years ago
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    I know but where's the external point ?

  36. perl
    • 3 years ago
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    |dw:1319452750463:dw|

  37. Mimi_x3
    • 3 years ago
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    Imaginary ?

  38. perl
    • 3 years ago
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    |dw:1319452790390:dw|

  39. Mimi_x3
    • 3 years ago
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    Is the parabola imaginery or somethinbg ?

  40. Mimi_x3
    • 3 years ago
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    imaginary* something*

  41. perl
    • 3 years ago
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    |dw:1319452894222:dw|

  42. Mimi_x3
    • 3 years ago
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    I don't get your diagrams ? sorry

  43. perl
    • 3 years ago
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    |dw:1319453015665:dw|

  44. Mimi_x3
    • 3 years ago
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    I don't get it ? >_<

  45. perl
    • 3 years ago
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    yeah the diagram is a beast

  46. Mimi_x3
    • 3 years ago
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    lols, why is there like 3 digrams ? i dont get it >_<

  47. perl
    • 3 years ago
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    |dw:1319453240562:dw|

  48. Mimi_x3
    • 3 years ago
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    lols, this is not 3D trig xD

  49. perl
    • 3 years ago
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    can you do better ?

  50. Mimi_x3
    • 3 years ago
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    Nope, i dont know how to start, the problem is where is T

  51. perl
    • 3 years ago
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    just a point below the parabola

  52. perl
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    like in that pdf

  53. Mimi_x3
    • 3 years ago
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    But it won't make sense, and i dont know how to prove

  54. perl
    • 3 years ago
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    have faith

  55. Mimi_x3
    • 3 years ago
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    have faith ?

  56. perl
    • 3 years ago
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    i think they mean the segment R T

  57. perl
    • 3 years ago
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    yes have faith when you draw it that it will all come into place, the problem will unravel

  58. perl
    • 3 years ago
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    i am atheist, what did you think i mean ?

  59. Mimi_x3
    • 3 years ago
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    idk, never mind then i will work it out myself

  60. perl
    • 3 years ago
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    |dw:1319453605908:dw|

  61. Mimi_x3
    • 3 years ago
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    ty (:

  62. perl
    • 3 years ago
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    im working on it

  63. perl
    • 3 years ago
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    would be nice if they gave you a diagram

  64. Mimi_x3
    • 3 years ago
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    Yeah i know, none of the hard questions have diagrams =/

  65. perl
    • 3 years ago
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    i think RT is the line segment though

  66. Mimi_x3
    • 3 years ago
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    hm idk

  67. perl
    • 3 years ago
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    |dw:1319454238070:dw|

  68. agdgdgdgwngo
    • 3 years ago
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    \[\text{An equation of the parabola with focus }(0,p)\text{ and directrix }y = -p\text{ is}\] \[x^2=4py\]

  69. agdgdgdgwngo
    • 3 years ago
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    so the directrix would be y = -a in your question :-D

  70. perl
    • 3 years ago
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    yes

  71. Mimi_x3
    • 3 years ago
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    Yeah, but im having problem proving

  72. agdgdgdgwngo
    • 3 years ago
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    If the chord of contact of the tangents to a parabola x² = 4ay, from an external point T(x1, y1) meets the directrix at R, prove that RT subtends a right angle at the focus. chord of contact of tangents to parabola is xx1=2a(y+y1) directrix is y=-a substitute into that, x=[2a(-a+y1)]/x1 y=-a R([2a(-a+y1)]/x1,-a) let S be the focus d/dx RS = 2a/ -[2a(-a+y1)]/x1= x1/(a-y1) d/dx TS=(y1-a)/x1 d/dx RS * d/dx TS= -1

  73. Mimi_x3
    • 3 years ago
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    lols, you got it from boredofstudies right ?

  74. agdgdgdgwngo
    • 3 years ago
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    yes... :-P

  75. perl
    • 3 years ago
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    how did you get chord of tangents equation

  76. perl
    • 3 years ago
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    hey, can you draw a diagram then of whats going on, please post the link

  77. Mimi_x3
    • 3 years ago
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    Yeah but the problem is that I don't understand the solution

  78. agdgdgdgwngo
    • 3 years ago
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    it helps to draw a diagram.

  79. perl
    • 3 years ago
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    can you post the link to the solution please , so i can read it

  80. agdgdgdgwngo
    • 3 years ago
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    http://community.boredofstudies.org/showthread.php?t=192825&page=1

  81. Mimi_x3
    • 3 years ago
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    But what's xx1 ? im trying to figure it out

  82. agdgdgdgwngo
    • 3 years ago
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    http://wiki.answers.com/Q/Find_the_equation_of_chord_of_contact_tangent_to_the_parabola

  83. Mimi_x3
    • 3 years ago
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    oh i get it, but where did y=1/2(p+q)x-apq come from ?

  84. agdgdgdgwngo
    • 3 years ago
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    xx1 must mean something like \[x_1\times x= 2a(y+y_1)\]

  85. Mimi_x3
    • 3 years ago
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    where did 2a come from ?

  86. agdgdgdgwngo
    • 3 years ago
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    good question

  87. agdgdgdgwngo
    • 3 years ago
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    \[parabola: x^2 = 4py\]\[a = \frac{1}{4p}\]

  88. Mimi_x3
    • 3 years ago
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    no, isn't it x^2 = 4ay y = x^2/4a y' = 2x/4a = x/2a

  89. perl
    • 3 years ago
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    |dw:1319455562515:dw|

  90. perl
    • 3 years ago
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    i wish there were other colors, but from T draw two tangents to the parabola

  91. agdgdgdgwngo
    • 3 years ago
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    perl's diagram works.... but how do we get there using calculus and derivatives if we don't even know the equation for chord of contact of the tangents to a parabola

  92. Mimi_x3
    • 3 years ago
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    oh the equation of the tangent to the parabola is y=px-ap^2 right ?

  93. perl
    • 3 years ago
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    |dw:1319455702848:dw|

  94. perl
    • 3 years ago
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    they are given here http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf

  95. perl
    • 3 years ago
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    the chords of tangents formula

  96. agdgdgdgwngo
    • 3 years ago
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    now it all makes sense

  97. perl
    • 3 years ago
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    |dw:1319455885258:dw|

  98. perl
    • 3 years ago
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    i dont get that parametrisation, is it because they want x^2 = 4ay ? , what is p

  99. Mimi_x3
    • 3 years ago
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    Doesn't it involve parameters as well P(2ap,ap^2)

  100. agdgdgdgwngo
    • 3 years ago
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    they want something like the quotient of the slopes between 2 lines is equal to -1

  101. agdgdgdgwngo
    • 3 years ago
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    and one of those lines should be RT

  102. perl
    • 3 years ago
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    |dw:1319456032235:dw|

  103. perl
    • 3 years ago
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    crap, it missed the R it is the right bottom corner

  104. Mimi_x3
    • 3 years ago
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    Im confused, is parameters involved ?

  105. agdgdgdgwngo
    • 3 years ago
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    no parameter seems to be involved :-(

  106. perl
    • 3 years ago
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    actually a is the parameter, you have to be careful, this word parameter is used in different senses

  107. perl
    • 3 years ago
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    , for instance in y = mx + b , m and b are parameters . now another sense of parameter is used when we 'parametrize' an equation. this means we introduce a third variable, often t

  108. Mimi_x3
    • 3 years ago
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    isn't parameters P(2ap,ap^2) So there must be a point for the parameters ?

  109. agdgdgdgwngo
    • 3 years ago
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    meh, I assumed m and b to be constant

  110. perl
    • 3 years ago
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    right, but they are considered 'parameters'

  111. Mimi_x3
    • 3 years ago
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    but its not a straight line so it cant be y=mx + b

  112. perl
    • 3 years ago
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    in the sense that they change your line . so for x^2 = 4ay, the parameter is a since that produces different parabolas by tweaking a

  113. perl
    • 3 years ago
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    think of in this sense a parameter is your tweaking constant

  114. agdgdgdgwngo
    • 3 years ago
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    :-(

  115. perl
    • 3 years ago
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    the parameter is 'a'*

  116. agdgdgdgwngo
    • 3 years ago
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    I thought a was just a slope

  117. Mimi_x3
    • 3 years ago
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    Im lost =/ Parameter P(2ap,ap^2)_

  118. perl
    • 3 years ago
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    changing 'a' changes your parabola. but for the line y = mx + b, changing m,b changes your line

  119. perl
    • 3 years ago
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    these are your tweaking values .

  120. perl
    • 3 years ago
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    I dont like their use of p, scratch that and use t instead

  121. perl
    • 3 years ago
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    ok they are using 'parametric equations' , x = f(t), y = g(t)

  122. perl
    • 3 years ago
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    so x = 2at, y = at^2

  123. agdgdgdgwngo
    • 3 years ago
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    I like to use matrices and vectors.

  124. Mimi_x3
    • 3 years ago
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    Yeah thats in cartesian form

  125. perl
    • 3 years ago
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    ok, use a matrix or vector, id like to see that

  126. perl
    • 3 years ago
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    mimi, when you remove t, you get cartesian equation

  127. Mimi_x3
    • 3 years ago
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    matrix or vector ?

  128. Mimi_x3
    • 3 years ago
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    Oh yeah my bad i forgot xD

  129. agdgdgdgwngo
    • 3 years ago
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    http://img392.imageshack.us/img392/8476/lach11n72fb.jpg

  130. Mimi_x3
    • 3 years ago
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    I haven't learnt that complicated stuff yet =/

  131. agdgdgdgwngo
    • 3 years ago
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    you will learn it soon.

  132. Mimi_x3
    • 3 years ago
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    Hope not looks hard already =/

  133. perl
    • 3 years ago
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    thats curve fitting, no big deal

  134. agdgdgdgwngo
    • 3 years ago
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    you will find it easier than conics

  135. Mimi_x3
    • 3 years ago
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    Whats conics ?

  136. perl
    • 3 years ago
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    its a quick way to fit a quadratic sequence, cubic, etc

  137. perl
    • 3 years ago
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    mimi, what we are doing is conics, oh dea

  138. agdgdgdgwngo
    • 3 years ago
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    conic sections: parabolas, circles, ellipses, hyperbolas, stuff.

  139. perl
    • 3 years ago
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    oh dear *

  140. agdgdgdgwngo
    • 3 years ago
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    you do them along parametric equations usually.

  141. perl
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    agdg, put the proper definition on board ;)

  142. agdgdgdgwngo
    • 3 years ago
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    well, they're what you get when you cut a cone.

  143. Mimi_x3
    • 3 years ago
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    LOL, this is not conics

  144. perl
    • 3 years ago
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    a conic has a specific algebraic equation in x and y , quadratic Ax^2 + Bxy + cY^2 + dy + ex + f = 0

  145. perl
    • 3 years ago
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    yes it is, a parabola is a conic

  146. perl
    • 3 years ago
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    a conic is a family of curves, as agdg pointed out

  147. Mimi_x3
    • 3 years ago
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    But Ax^2 + Bxy + cY^2 + dy + ex + f = 0 ?

  148. agdgdgdgwngo
    • 3 years ago
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    forget the algebra and xs and ys and f(x)s, you will understand them really well once you picture them geometrically

  149. agdgdgdgwngo
    • 3 years ago
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    http://mathworld.wolfram.com/ConicSection.html

  150. perl
    • 3 years ago
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    right

  151. Mimi_x3
    • 3 years ago
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    Um, this doesnt look like a conic

  152. perl
    • 3 years ago
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    agdg, whats the nth term of this quadratic sequence 3,5,10,18...

  153. perl
    • 3 years ago
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    lets test your matrix skills

  154. agdgdgdgwngo
    • 3 years ago
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    hmmm... is it infinity?

  155. perl
    • 3 years ago
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    no, the nth term

  156. perl
    • 3 years ago
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    give me a formula for the sequence

  157. perl
    • 3 years ago
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    obviously it diverges, ;)

  158. agdgdgdgwngo
    • 3 years ago
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    its \[2^n + 2^\frac{n}{2}\]

  159. perl
    • 3 years ago
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    no, lol

  160. agdgdgdgwngo
    • 3 years ago
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    wait! it's \[\frac{2^n+2^{\frac{n}{2}}}{2}\]

  161. perl
    • 3 years ago
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    its a quadratic sequence, like in your matrix

  162. agdgdgdgwngo
    • 3 years ago
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    how do we find it :-D

  163. agdgdgdgwngo
    • 3 years ago
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    so what's the 5th term

  164. perl
    • 3 years ago
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    ok so first we find the differences

  165. perl
    • 3 years ago
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    the first row of differences is 2 , 5, 8, 11 , ... the second row of differences is 3, 3, 3, ... so you know its a quadratic sequence

  166. agdgdgdgwngo
    • 3 years ago
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    that looks like a long matrix

  167. perl
    • 3 years ago
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    3, 5, 10, 18, 29 ... thats the sequence

  168. perl
    • 3 years ago
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    it should fit f(n) = an^2 + bn + c

  169. agdgdgdgwngo
    • 3 years ago
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    I guess that works.

  170. agdgdgdgwngo
    • 3 years ago
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    how do we multiply matrices?

  171. perl
    • 3 years ago
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    are you asking in general ?

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