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Quadric surfaces problem please help.

  1. Sep 15, 2009 #1
    Here is the problem exactly how it is written on my paper...

    Consider the surfaces x^2+2y^2-z^2+3x=1 and 2x^2+4y^2-2z^2-5y=0.
    a. What is the name of each surface?
    b. Find an equation for the plane which contains the intersection of these two surfaces.

    That is the question. For "a", I generally know what needs to be done, but I can't figure out how to get the equations into the form of a standard quadric surface equation. Do I just complete the square, or what? Also, I really have no idea how to do "b". Can someone solve this and please explain to me how to do it? Thanks in advance.
  2. jcsd
  3. Sep 15, 2009 #2


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    This sounds like a reasonable place to start -- even if it doesn't work, it's still something to try. What'd you get after you did it? Why'd you abandon that line of attack?
  4. Sep 15, 2009 #3
    I got [tex]\frac{(x+3/2)^2}{13/4} + \frac{y^2}{13/8} - \frac{z^2}{13/4} = 1[/tex] and [tex]\frac{x^2}{25/128} + \frac{(y-5/8)^2}{25/256} - \frac{z^2}{25/128} = 1[/tex]

    Based on these equations it seems like they are both hyperboloid of one sheet surfaces. Is this correct? These answers seem a bit strange, which is why I made the thread on here to ask about it.

    I really don't know what to do about "b" at all, I really need help with that.
  5. Sep 15, 2009 #4


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    I haven't multiplied it out, but those equations look roughly like I would expect.

    For "b", it seems the obvious first thing to try is to look at the intersection of the two surfaces! It's the set of solutions to both equations, so you should be able to do some simplification.

    Another idea that springs to mind is to try and think of some kind of information that would let you determine the plane -- and then try and compute that information.
  6. Sep 15, 2009 #5
    Sorry, but I am still unsure of how to start. How should I equate the two equations together to find the points of intersection? Can you start it for me and just show me what you would do? Thanks.
  7. Sep 16, 2009 #6


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    Yes, since both expressions are equal to 1, they are equal to each other. Although, instead of the rather complicated expressions you derived, I would recommend going back to [itex]x^2+ 2y^2- z^2+ 3x- 1= 0= 2x^2+ 4y^2- 2z^2- 5y. You can solve that for, say, [itex]z^2[/itex] in terms of x and y and put that back into the original equations to get an equation is x and y. Then solve for either x or y in terms of the other and, perhaps, use the remaining variable as parameter.
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