1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relationships of hyperbolic Paraboloids

  1. Oct 6, 2012 #1
    Hey everyone, I was wondering what you could tell me about the relationship between hyperbolic paraboloids. I have listed a set of 3 equations and was wondering what I can do with them? Can I solve for z, can I get the intersection of the equations? Can I get generalized solution of any kind? What does all three equations existing in the same domain mean (what would it look like to plot all of them in 3d, what information would it garner?).

    The coefficents are irrelevant, I am wondering about the type of equation, listed below is an example set.

    7*x*y +15y +11x + 4 = z
    7*x*y +26y +57x + 27 = z
    7*x*y +72y +84x + 81 = z

    Could you just throw some ideas at me, please.

    Thank you!!
  2. jcsd
  3. Oct 6, 2012 #2


    User Avatar
    Science Advisor

    You can easily eliminate two variables to get a quadratic in the third. Solve and go back. You will have two possible solutions, so you need to verify if both are valid.
  4. Oct 6, 2012 #3
    I am not sure how to handle the 7*x*y in order to get the quadratic. I tried using the solver applets in Wolfram alpha and it is having issues with it as well.

    Thank you for the quick response!
  5. Oct 7, 2012 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Not sure how general are the cases you're interested in. In your example the only quadratic term anywhere is xy. So you can eliminate z between two eqns to obtain an expression for xy in terms of linear form on x and y. Then use that to eliminate xy from two eqns (one of the first two and the third) to get eqns of two intersecting planes, and hence of a line. Use those to eliminate y and z (say) from any one of the three eqns to get a quadratic.
  6. Oct 7, 2012 #5


    User Avatar
    Science Advisor

    Start with a(i)xy + b(i)x + c(i)y + d(i) = z, i = 1,2,3

    Eliminate z to get two equations:
    (a(1)-a(j))xy + (b(1)-b(j))x + (c(1)-c(j))y + d(1)-d(j) = 0, j =2,3

    Solve the first equation for x:

    x = {d(2) - d(1) + (c(2) - c(1))y}/{b(1) - b(2) + (a(1) - a(2))y}

    Substitute for x in the second equation and clear the denominator. The result is a quadratic in y.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook