Quadratic in two variables

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Homework Statement


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

Homework Equations




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.
 

Answers and Replies

  • #2
Simon Bridge
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I'd have started by considering what sort of objects the LHS and the RHS are ...
 
  • #3
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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?
 
  • #4
RUber
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What if you set the whole thing equal to zero and found solutions?
 
  • #5
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Why would I want to do that? I can't just arbitrarily assign something a value.
 
  • #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.
 
  • #7
Simon Bridge
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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?
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?
 
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  • #8
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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
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Solve for x in terms of y, and you'll see the answer immediately.

Chet
How do I deal with the xy term when trying to solve for x?
 
  • #11
PeroK
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How do I deal with the xy term when trying to solve for x?

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}##
 
  • #12
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How do I deal with the xy term when trying to solve for x?
You're solving for x in terms of y, so it's an ordinary quadratic in x. Use the quadratic formula.

Chet
 
  • #13
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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
 
  • #14
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Btw, finally solving this question made my day :)
 

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