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Surds & Length of straight line

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data
    1)The normal to C ( 3y = x + 20 ) at P (4, 8) cuts the x-axis at the point Q.
    Find the length PQ, giving your answers in a simplified surd form.

    2) write [tex] \dfrac{2\sqrt{x} + 3}{x} [/tex] in the form [tex] 2x^p + 3x^q [/tex] where p and q are constants

    2. Relevant equations

    y = mx + c? I'm not sure


    3. The attempt at a solution

    for 1) i let y = 0 and got Q (-20,0) but i don't know how to find the length of the two points.

    for 2)
    [tex] \dfrac{3}{x} = 3x^{-1} [/tex] but i don't know what [tex]\dfrac{2\sqrt{x}}{x} [/tex] is.

    thanks
     
    Last edited: Nov 5, 2011
  2. jcsd
  3. Nov 5, 2011 #2

    LCKurtz

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    Q is the point where the normal intersects the x axis. You have to start by writing the equation of the normal line. And you don't find the "length of two points". You find the distance between the two points.
     
  4. Nov 5, 2011 #3

    LCKurtz

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    Write it with exponents and use the rules of exponents to simplify it.
     
  5. Nov 5, 2011 #4
    Alright, for 1)

    I let y = 0 and got Q as (-20,0) (seeing as it says it crosses the x-axis)

    so [tex]
    \sqrt{(4-(-20))^2 + (8-0)^2}
    = \sqrt{640}
    = 8\sqrt{10}
    [/tex]
    for 2)

    [tex] \dfrac{2\sqrt{x}}{x} = 2x^{-\dfrac{1}{2}}

    = 2x^{-\dfrac{1}{2}} + 3x^{-1}

    [/tex]

    Have i gone wrong anywhere?
     
  6. Nov 5, 2011 #5

    LCKurtz

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    Did you even read my post?? Q = (-20,0) has nothing to do with the problem.
    You have worked the exponents correctly, but that last line has = signs between things that aren't equal.
     
  7. Nov 5, 2011 #6
    I wrote = just to show it goes to that answer, but whatever i understand that question now, thanks.

    For 1) the question says that the normal to C at P is 3y = x + 20 (its on the question before that), P is (4,8) also on the question before that.

    It says that the normal at (4,8) cuts the x-axis at Q, to find where it crosses the x axis you would have to let y = 0 no? If not i'm completely mistaken as thats the only way i'm seeing the question at the moment, i don't see what your trying to say.

    thanks again.
     
  8. Nov 5, 2011 #7

    LCKurtz

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    We aren't mind readers here. The statement I have bolded is readily interpreted "The normal to [the curve] C [whose equation is] 3y = x + 20 ... meaning you have this curve C whose equation is given and the first problem is to find its normal. Why do you even mention it is normal to some curve C which isn't given and is irrelevant? How are we supposed to know what is on "the question before that"?
     
  9. Nov 5, 2011 #8
    I thought it was pretty clear that the NORMAL TO C (3y=x+20), i didn't want to mention other questions because i've already solved the other questions and mentioned the data which was needed for this question...

    thanks for your help anyway, appreciated.
     
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