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Math Help - Quadratic in 2 variables (graphs)

  1. #1
    r45
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    Quadratic in 2 variables (graphs)

    Hi,

    I have been looking at quadratics of the form:

    ax^2 + bxy + cy^2 + dx + ey + f = 0

    From this I determined that:

    y = \frac{-bx - e \pm \sqrt{(b^2 - 4ac)x^2 + (2be - 4cd)x + (e^2 - 4cf)}}{2c}

    In particular I am looking at xy - 4x - 2y = 0 which is in standard form with:

    a = 0, b = 1, c = 0, d = -4, e = -2, f = 0

    However the above formula fails since the denominator would be 0. So I approached differently:

    xy - 2y = 4x
    y(x - 2) = 4x
    y = \frac{4x}{x - 2}

    So I used a graphing program to sketch this:



    I am asked to "identify the type of graph obtained" - what sort of graph is this? I would say hyperbola but I don't think that's necessarily right?
    Last edited by r45; March 27th 2010 at 08:57 AM.
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  2. #2
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    Quote Originally Posted by r45 View Post
    Hi,

    I have been looking at quadratics of the form:

    ax^2 + bxy + cx^2 + dx + ey + f = 0

    From this I determined that:

    y = \frac{-bx - e \pm \sqrt{(b^2 - 4ac)x^2 + (2be - 4cd)x + (e^2 - 4cf)}}{2c}

    In particular I am looking at xy - 4x - 2y = 0 which is in standard form with:

    a = 0, b = 1, c = 0, d = -4, e = -2, f = 0

    However the above formula fails since the denominator would be 0. So I approached differently:

    xy - 2y = 4x
    y(x - 2) = 4x
    y = \frac{4x}{x - 2}

    So I used a graphing program to sketch this:



    I am asked to "identify the type of graph obtained" - what sort of graph is this? I would say hyperbola but I don't think that's necessarily right?
    you are correct ... it is a hyperbola.
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  3. #3
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    Quote Originally Posted by r45 View Post
    ax^2 + bxy + cx^2 + dx + ey + f = 0

    From this I determined that:

    y = \frac{-bx - e \pm \sqrt{(b^2 - 4ac)x^2 + (2be - 4cd)x + (e^2 - 4cf)}}{2c}
    Huh? There's no y^2 in original equation
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  4. #4
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    Yes, it's a hyperbola. Any quadratic in two variables is a "conic section" (provided you include "degenerate" forms). Since this "looks like" a hyperbola and the only conic section that looks like that is the hyperbola, it is a hyperbola! (Obviously that vertical line between the two branches is an artifact of the graphing program.) Note that xy= 1 has no x^2 or y^2 term but is a hyperbola.

    More specifiically, if you let x= x'cos(\theta)- y'sin(\theta) and y= x'sin(\theta)+ y' cos(\theta) then xy= x'^2 cos(\theta)sin(\theta)+ x'y'(cos^2(\theta)- sin^2(\theta)- y'sin^2(\theta). Now if \theta= \pi/4, sin(\theta)= cos(\theta)= \sqrt{2}/2 so that cos^2(\theta)- sin^2(\theta)= 0 and xy= x'^2/\sqrt{2}- y'^2/\sqrt(2).

    Then xy- 4x- 2y= x'^2/\sqrt{2}- y'^2/\sqrt{2}- 6x'/\sqrt{2}+ 2y'/\sqrt{2}.

    You could now determine the center of the hyperbola by completing the square. Also, a "standard" hyperbola, of the form \frac{x^2}{a^2}- \frac{y^2}{b^2}= 1 has axes of symmetry along the x and y axes. Here it has been rotated \pi/4 radians or 45 degrees.


    About those "degenerate cases": Obviously, the quadratic equation x^2+ y^2= 0 is satisfied only by the single point (0, 0) while the set of points satisfying x^2+ y^2= -1 is the empty set. We can think of those as "degenerate" circles. x^2- y^2= 0 can be written as (x- y)(x+ y)= 0 so will be satisfied by (x, y) so that either y= x or y= -x, a pair of lines intersecting at the origin. We can think of that as a "degenerate" hyperbola. Finally, x^2- 2xy+ y^2= 0 is the same as (x- y)^2= 0 whose graph is the single line y= x. We can think of that as a "degenerate" parabola.
    Last edited by HallsofIvy; March 27th 2010 at 10:52 AM.
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  5. #5
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    Quote Originally Posted by r45 View Post
    Hi,

    I have been looking at quadratics of the form:

    ax^2 + bxy + c\textcolor{red}{y^2} + dx + ey + f = 0
    typo
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  6. #6
    r45
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    Right thanks a lot for the responses.

    So in general, if we have:

    y = \frac{ax}{x-b}

    This would represent a "standard" hyperbola except rotated anti-clockwise by \frac{\pi}{a} radians, and translated by b units on the y-axis?
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  7. #7
    r45
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    Or would that be translated by b^2 units?
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    Quote Originally Posted by r45 View Post
    Right thanks a lot for the responses.

    So in general, if we have:

    y = \frac{ax}{x-b}

    This would represent a "standard" hyperbola except rotated anti-clockwise by \frac{\pi}{a} radians, and translated by b units on the y-axis?
    Quote Originally Posted by r45 View Post
    Or would that be translated by b^2 units?
    Neither. That becomes xy- by= ax or xy- ax- by= 0. That will be a hyperbola rotated through \pi/4 radians. The angle of rotation is determined solely by the quadratic terms as I showed before.

    It would be translated horizontally by b/2 and vertically by a/2.
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  9. #9
    r45
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    Ah OK right I just re-read through your earlier post and I understand now.

    I also have a follow up question:

    If f(x,y) = ax^2 + bxy + cy^2 + dx + ey + f can be factored into a product of degree one factors, what can we say about the graph of f(x,y) = 0?
    Last edited by r45; March 27th 2010 at 11:50 AM.
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  10. #10
    r45
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    There are a number of self-evident points we could make, like the number of distinct factors would equal the total number of intersections with the x and y axes. But I have a feeling the question is seeking something less obvious?
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    Quote Originally Posted by r45 View Post
    Ah OK right I just re-read through your earlier post and I understand now.

    I also have a follow up question:

    If f(x,y) = ax^2 + bxy + cx^2 + dx + ey + f can be factored into a product of degree one factors,
    what can we say about the graph of f(x,y) = 0?
    WHERE is y^2 ?????????????
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  12. #12
    r45
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    Ahh sorry I'm hopeless! I've edited it to show the correct term now.
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  13. #13
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    Since it is quadratic, saying f(x) can be factored into "a product of degree one factors" means it is either f(x,y)= (ax+ by)(cx+ dy)= 0 or (ax+ by)^2= 0.

    In the first case, the graph is the two straight lines ax+ by= 0 and cx+ dy= 0, which I referred to before as a "degenerate hyperbola". In the second case, the graph is the single straight line, ax+ by= 0, which is a "degenerate parabola".
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  14. #14
    r45
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    So if d, e or f are non-zero then the quadratic f(x,y) = ax^2 + bxy + cy^2 + dx + ey + f cannot be factored into a product of degree one factors?
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