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Prove parabolas intersect at right angles (polar eq'ns)

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the parabolas r=c/(1+cosθ)and r'=d/(1-cosθ) intersect at right angles.

    3. The attempt at a solution

    I found the points of intersection by setting the two equations equal, to which I got:
    cosθ = (c- d)/(c+d)
    θ = cos^-1[(c- d)/(c+d)]

    then i tried to find the slope of the two equations:

    x=dcosθ/1-cosθ ; y=dsinθ/1-cosθ

    dy/dθ = [dcosθ(1-cosθ)-(sinθ)dsinθ] / (1-cosθ)^2 = d(cosθ-1)/(1-cosθ)^2
    dx/dθ= [-dsinθ(1-cosθ)-(sinθ)dcosθ] / (1-cosθ)^2 = -dsinθ/(1-cosθ)^2

    dy/dx=cosθ-1/-sinθ

    x=ccosθ/1+cosθ ; y=csinθ/1+cosθ

    dy'/dθ = [-csinθ(1+cosθ)-(-sinθ)ccosθ] / (1+cosθ)^2 = -csinθ/(1+cosθ)^2
    dx'/dθ= [ccosθ(1+cosθ)-(-sinθ)csinθ] / (1+cosθ)^2 = c(cosθ+1)/(1+cosθ)^2

    dy/dx=cosθ+1/-sinθ

    Then I don't know what to do
     
  2. jcsd
  3. Feb 11, 2009 #2

    Dick

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    Two slopes are perpendicular if their product is -1. Are they? Use a trig identity.
     
  4. Feb 11, 2009 #3
    thanks!
     
  5. Feb 11, 2009 #4

    Dick

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    You don't have to assume that. sin(x)^2+cos(x)^2=1. So cos(x)^2-1=-sin(x)^2. ALWAYS. It's an identity.
     
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