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Intersection of a paraboloid and a plane

  1. Feb 19, 2006 #1
    Question: Consider the intersection of the paraboloid [tex]z = x^2 + y^2[/tex] with the plane [tex]x - 2y = 0[/tex]. Find a parametrization of the curve of intersection and verify that it lies in each surface.

    How I went about it:

    [tex]x = 2y[/tex]
    [tex]z = (2y)^2 + y^2 = 5y^2[/tex]

    Set [tex]y = t[/tex], then

    [tex]x = 2t[/tex]
    [tex]y = t[/tex]
    [tex]z = 5t^2[/tex]

    I don't know that my answer is wrong, I'm just not certain if I am going about it the correct way. If someone could let me know I would appreciate it. Thanks.

  2. jcsd
  3. Feb 19, 2006 #2


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    Homework Helper

    I plotted both the parabolaoid and the plane in one plot and the parametric spacecurve in another plot (using Maple) and they do appear to agree, also your work is without flaw. Good job.
  4. Feb 19, 2006 #3


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    Science Advisor

    Yes, it really is that simple!
  5. Feb 19, 2006 #4
    Thanks for the replys.

  6. Feb 19, 2006 #5
    If only the rest of calculus were this easy :smile:
  7. Feb 22, 2009 #6
    A similar question - but more variables are present:

    Find the intersection of the plane 2x-y-3z=15 and the paraboloid 3z=(x^2)/16 + (y^2)/9

    I was unable to use the above method because of the "z" variable in the plane equation. Can I still use the above method? Or is something else necessary?

    I plugged 3z=15+2x-y into the equation for the paraboloid and started getting imaginary numbers, was that correct?

    "At the end of the number line is a rainbow made out of nothing but primes..."
    Last edited: Feb 22, 2009
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