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Quadratic in two variables

  1. Oct 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Given x^2 - xy + y^2 = 4(x+y-4) where x, y both are real then the number of pairs (x, y) satisfying the equation will be

    (A) Only one
    (B) Two
    (C) Three
    (D) No pair

    2. Relevant equations


    3. The attempt at a solution
    The equation can be written as (x-2)^2 + (y-2)^2 + 8 = xy on opening the brackets and completing the square.
    Beyond this, I really don't know what to do. Hit and trial doesn't seem to work very well either.
     
  2. jcsd
  3. Oct 29, 2014 #2

    Simon Bridge

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    I'd have started by considering what sort of objects the LHS and the RHS are ...
     
  4. Oct 29, 2014 #3
    One side is a hyperbola, the other is a line. So, we are talking about the intersection of a line with a hyperbola. What beyond that?
     
  5. Oct 29, 2014 #4

    RUber

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    What if you set the whole thing equal to zero and found solutions?
     
  6. Oct 29, 2014 #5
    Why would I want to do that? I can't just arbitrarily assign something a value.
     
  7. Oct 29, 2014 #6

    RUber

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    The way I drew this one out was using your expansion ##(x-2)^2+(y-2)^2 = R^2 = xy - 8 ##. Where ##R^2## is the radius of the circle. Drawing the two function as you vary ##z=R^2## makes some telling plots.

    Another method would be to let ##f(x,y)=(x-2)^2+(y-2)^2-xy+8## and use the derivative to find a set of critical points. These may also be helpful in determining when this relation could be true.
     
  8. Oct 30, 2014 #7

    Simon Bridge

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    Well you need to know when one side is equal to the other side ... note: how do you figure a line and a hyperbola? What happens when you plot z(x,y) for just one side?

    Note: putting the relation in form f(x,y)=0 gives you a family of curves.

    What we are doing here is trying to jog your recollection of something in your coursework.
    Have you not just been learning about techniques for dealing with this sort of thing?
     
    Last edited: Oct 30, 2014
  9. Oct 31, 2014 #8
    If this were a standard hyperbola, I could probably figure out the points of intersection between a line and a hyperbola. However, this is not a standard hyperbola. So, I don't know.
     
  10. Oct 31, 2014 #9
    Solve for x in terms of y, and you'll see the answer immediately.

    Chet
     
  11. Oct 31, 2014 #10
    How do I deal with the xy term when trying to solve for x?
     
  12. Oct 31, 2014 #11

    PeroK

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    You're missing something that perhaps not a lot of people notice. Suppose you have a quadratic equation:

    ##x^2 + bx + c = 0## with solution ##x= \frac{-b \pm \sqrt{b^2-4c}}{2}##

    Well, you can look at b & c as variables. And you could re-interpret this equation as:

    ##f(x, b, c) = x^2 + bx + c = 0##

    You are reluctant to treat "y" as a "constant", because y is a "variable". But, maybe there's no so much difference. So, if you replace b & c with y and z you get:

    ##f(x, y, z) = x^2 + xy + z = 0## and this still has solutions ##x= \frac{-y \pm \sqrt{y^2-4z}}{2}##
     
  13. Oct 31, 2014 #12
    You're solving for x in terms of y, so it's an ordinary quadratic in x. Use the quadratic formula.

    Chet
     
  14. Oct 31, 2014 #13
    On treating y as a 'constant' and simplifying, I get
    x= { y+4 ± √(3(y-4)(4-y)) }/ 2 ----- (1)
    (y-4)(4-y) ≥ 0
    -(y-4)^2 ≥ 0
    (y-4)^2 ≤ 0
    Since a square can only be zero or positive,
    y-4 = 0
    So, y=4.

    So, the entire root part of (1) goes out.
    So, this has only one solution.
    Thanks everyone :D
     
  15. Oct 31, 2014 #14
    Btw, finally solving this question made my day :)
     
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