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Does this curve exist?

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  • 23-08-2015 10:47am
    #1
    Closed Accounts Posts: 895 ✭✭✭


    Hi,

    I would like to know does this curve exist and if so what form does it take:

    Take any points on a curve, multiply the (absolute) value of x and y and the answer is the same for any value of x,y.

    So for example, if points (3,4) and (4,3) are on a curve, 3*4 = 4*3 = 12, points (6,2) and (2,6) might be on the curve too because 6*2 = 2*6 = 12. Hence the curve has to have points all of which are true for XY=12?

    Thanks!


Comments

  • Registered Users Posts: 5,619 ✭✭✭TheBody


    If I understand your question correctly, there are two obvious ones, the lines y=0 and the line x=0. (i.e the x any y axis respectively)

    For example, the any point along the x-axis is of the form (x,0). So when we multiply the x and y coordinates we always get 0.

    Edit: I just noticed you want x and y to vary. If that is the case, I don't think there is a curve.


  • Closed Accounts Posts: 895 ✭✭✭Dughorm


    TheBody wrote: »
    If I understand your question correctly, there are two obvious ones, the lines y=0 and the line x=0. (i.e the x any y axis respectively)

    For example, the any point along the x-axis is of the form (x,0). So when we multiply the x and y coordinates we always get 0.

    Edit: I just noticed you want x and y to vary. If that is the case, I don't think there is a curve.

    Thanks The Body!


  • Registered Users Posts: 1,169 ✭✭✭dlouth15


    I might be misunderstanding the question but I think what you are asking is what is the curve or curves for the implicit equation

    [latex]\displaystyle{\left|xy\right|=\left|x\right|\left|y\right|=12}[/latex].

    It looks like there are two curves for which the above equation is true.

    [latex]\displaystyle{y=\pm12/x}[/latex]

    Here's a plot of these two curves.

    359686.png
    One curve is in blue, the other in orange.


  • Registered Users Posts: 5,619 ✭✭✭TheBody


    dlouth15 wrote: »
    I might be misunderstanding the question but I think what you are asking is what is the curve or curves for the implicit equation

    [latex]\displaystyle{\left|xy\right|=\left|x\right|\left|y\right|=12}[/latex].

    It looks like there are two curves for which the above equation is true.

    [latex]\displaystyle{y=\pm12/x}[/latex]

    Here's a plot of these two curves.

    plot2.svg?_subject_uid=275419639&w=AAAlvZa7T9gqZkmFkkgyb07CTsIESyYzOM8_2bDzd0iHcg&dl=1
    One curve is in blue, the other in orange.

    I think I may have misinterpreted the original question. What you have seems more like what they are asking.


  • Closed Accounts Posts: 895 ✭✭✭Dughorm


    dlouth15 wrote: »
    I might be misunderstanding the question but I think what you are asking is what is the curve or curves for the implicit equation
    .

    Bingo! That's exactly what I was looking for - is it true that if you draw a 45 degree angle from the origin that you will cut the curve in half and the point of intersection will be the point where x=y which is also the square root of, say 12 in a y=12/x curve?


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  • Registered Users Posts: 1,169 ✭✭✭dlouth15


    Dughorm wrote: »
    Bingo! That's exactly what I was looking for - is it true that if you draw a 45 degree angle from the origin that you will cut the curve in half and the point of intersection will be the point where x=y which is also the square root of, say 12 in a y=12/x curve?
    Yes, at the intersection you will have the point [latex]x=\sqrt{12},\,y=\sqrt{12}[/latex] and also [latex]x=-\sqrt{12},\,y=-\sqrt{12}[/latex].

    This is because if you make y=x, then y=12/y and therefore y^2=12. Since x and y are the same, x^2 = 12 too.


  • Closed Accounts Posts: 895 ✭✭✭Dughorm


    dlouth15 wrote: »
    Yes, at the intersection you will have the point [latex]x=\sqrt{12},\,y=\sqrt{12}[/latex] and also [latex]x=-\sqrt{12},\,y=-\sqrt{12}[/latex].

    This is because if you make y=x, then y=12/y and therefore y^2=12. Since x and y are the same, x^2 = 12 too.

    Thanks again dlouth - this is actually really intuitive...it's great to marry the graphical and the algebraic reasoning like you have done there!


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