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Parabolas that intersect, tricky one

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data
    There are two parabolas which intersect.

    One is y = x^2
    The other is the same size, but is on a rotated cartesian plane, 45º CW of the first parabola.
    Find the the highest intersection point.
    [​IMG]
    2. Relevant equations
    Anything relevent.


    3. The attempt at a solution
    First of all, I see that the fact there is an angle being used. So I think we'd need to get the parabolas in another co-ordinate system based on angles. Polar co-ordinates are there for us.

    So I construct a triangle with x, y, r and theta. I need to get r in terms of theta.

    For one, tan0 = y/x
    y = x^2, as it's a parabola.

    So

    [tex]
    tan\theta = x^2 / x,

    tan\theta = x,

    r = \sqrt{x^2 + y^2},


    r = \sqrt{x^2 + (x^2)^2},


    r = \sqrt{x^2 + x^4},

    r = \sqrt{tan^2\theta + tan^4\theta}[/tex]

    So by changing the theta, I could rotate the parabola.
    By adding 45º, (pi / 4), it would rotate it to the correct position.

    So I need to solve:
    [tex]
    \sqrt{tan^2\theta + tan^4\theta} = \sqrt(tan^2(\theta + \pi/4) + tan^4(\theta + \pi/4))[/tex]

    Here I get stuck. I can't seem to solve this.
     
  2. jcsd
  3. Nov 18, 2007 #2
    I used SOLVE on Mathematica 6; I got these 10 solutions. I guessed that on the right memeber of your equation the squared rooth is "covering" all the member.

    {{\[Theta] -> -ArcCos[-1/2 Sqrt[2 - Sqrt[2]]]}, {\[Theta] ->
    ArcCos[Sqrt[2 - Sqrt[2]]/2]}, {\[Theta] ->
    ArcCos[-1/2 Sqrt[2 + Sqrt[2]]]}, {\[Theta] -> -ArcCos[Sqrt[
    2 + Sqrt[2]]/
    2]}, {\[Theta] -> -ArcCos[-1/2 Sqrt[
    3 - 2 Sqrt[2] - \[ImaginaryI] Sqrt[
    7 - 4 Sqrt[2]]]]}, {\[Theta] ->
    ArcCos[1/2 Sqrt[
    3 - 2 Sqrt[2] - \[ImaginaryI] Sqrt[
    7 - 4 Sqrt[2]]]]}, {\[Theta] -> -ArcCos[-1/2 Sqrt[
    3 - 2 Sqrt[2] + \[ImaginaryI] Sqrt[
    7 - 4 Sqrt[2]]]]}, {\[Theta] ->
    ArcCos[1/2 Sqrt[
    3 - 2 Sqrt[2] + \[ImaginaryI] Sqrt[
    7 - 4 Sqrt[2]]]]}, {\[Theta] ->
    ArcCos[-Sqrt[
    3/4 + 1/Sqrt[2] -
    1/4 \[ImaginaryI] Sqrt[7 + 4 Sqrt[2]]]]}, {\[Theta] -> -ArcCos[
    Sqrt[3/4 + 1/Sqrt[2] -
    1/4 \[ImaginaryI] Sqrt[7 + 4 Sqrt[2]]]]}, {\[Theta] ->
    ArcCos[-Sqrt[
    3/4 + 1/Sqrt[2] +
    1/4 \[ImaginaryI] Sqrt[7 + 4 Sqrt[2]]]]}, {\[Theta] -> -ArcCos[
    Sqrt[3/4 + 1/Sqrt[2] + 1/4 \[ImaginaryI] Sqrt[7 + 4 Sqrt[2]]]]}}
     
  4. Nov 18, 2007 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The second parabola is symmetric about the 45 degree angle line: the line y= x. The first is symmetric about x= 0 at 90 degrees from the x-axis. Their intersection must be on the line half way between 45 degrees and 90 degrees: 67.5 degrees from the x-axis. tan(67.5)= 1+ [itex]\sqrt{2}[/itex] so that is the line y= (1+[itex]\sqrt{2}[/itex])x. What are the points of intersection of that line with y= x2?
     
    Last edited: Nov 18, 2007
  5. Nov 20, 2007 #4
    Wow, HallsofIvy's answer is so simple and brilliant. I got the same answer as well, through solving that big tan equation. It left me with two positive angles, 7pi/8 and 3pi/8. I found the cartesian co-ordinates of the intersection, and noticed the decimal as .414, which is from root 2. After a little messing around, I saw the x-value was simply 1 + root2. Then, after a bit of messing with the y-co-ordinate, I saw that it was (1 + root2)^2. It has to be, since it's on the parabola y = x^2.

    Thank you all for your help. I now feel reasonably able to solve this sort of problem if I ever have to. Using either symmetry or using polar algebra. :)
     
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