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Thread: what's in a name?

  1. #1
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    what's in a name?

    Hi folks,

    I was looking at the following equation:

    $2x^2 + 3y^2 -4xy + 2x -6y + 3 = 0$

    I put it into a graphing tool and it looks like an ellipse. But is it an ellipse? I am saying it isn't because it has the 'xy' term in it.

    So my question is: how do I know what a shape is? The eye is not a precise instrument, so a thing can look like a circle and not be. My equation looks like an ellipse, but I'm guessing it isn't.

    So there are lots of shapes that look like ellipses and hyperbolas, but may not be. So what do we call these shapes?

    Any comments on this would be appreciated.
    Last edited by s_ingram; May 15th 2018 at 08:11 AM.
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  2. #2
    Member Walagaster's Avatar
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    Re: what's in a name?

    Quote Originally Posted by s_ingram View Post
    Hi folks,

    I was looking at the following equation:

    $2x^2 + 3y^2 -4xy + 2x -6y + 3 = 0$

    I put it into a graphing tool and it looks like an ellipse. But is it an ellipse? I am saying it isn't because it has the 'xy' term in it.

    So my question is: how do I know what a shape is? The eye is not a precise instrument, so a thing can look like a circle and not be. My equation looks like an ellipse, but I'm guessing it isn't.

    So there are lots of shapes that look like ellipses and hyperbolas, but may not be. So what do we call these shapes?

    Any comments on this would be appreciated.
    It is an ellipse. You can read about rotated conics and their discriminant here:
    Quadratic Curve Discriminant -- from Wolfram MathWorld
    Thanks from s_ingram
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  3. #3
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    Re: what's in a name?

    Hey Walagaster: can I borrow your cup?
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  4. #4
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    Re: what's in a name?

    With that "xy" term, try rotating the coordinate system. If you rotate the coordinate system through angle $\theta$, the old "x, y" coordinates are given in terms of the new "x', y'" coordinate system by $x= x'cos(\theta)- y'sin(\theta)$, $y= x'sin(\theta)+ y'cos(\theta)$. Multiplying those, $x^2= x'cos^2(\theta)- 2x'y'sin(\theta)cos(\theta)+ y'^2 sin^2(\theta)$, $xy= x'^2sin(\theta)cos(\theta)- x'y'(cos^2(\theta)- sin^2(\theta))- y' sin(\theta)cos(\theta)$, and $y^2= x'^2sin^2(\theta)+ 2x'y'sin(\theta)cos(\theta)+ y'^2cos^2(\theta)$.

    Putting those into $2x^2- 4xy+ 3y^2$, but looking only at the "x'y'" terms, we have $(-4 sin(\theta)cos(\theta)+ 8(cos^2(\theta)- sin^2(\theta))+ 6sin(\theta)cos(\theta))x'y'= (8cos^2(\theta)+ 2sin(\theta)cos(\theta)- 8sin^2(\theta))x'y'$. To eliminate that term we must have $8cos^2(\theta)+ 2sin(\theta)cos(\theta)- 8sin^2(\theta)= 0$ We can think of that as a quadratic equation for $cos(\theta)$ in terms of $sin(\theta)$ and solve it using the quadratic formula.

    A computationally simpler, but more "sophisticated" method is to write the quadratic part as a matrix multiplication: $\begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}2 & -2 \\ -2 & 3\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}$. I have allocated half of the "-4xy" to each of the first and second rows so this is a symmetric matrix and has a "full set" of eigenvectors. The eigenvalue equation is $\left|\begin{array}{cc}2- \lambda & -2 \\ -2 & 3- \lambda\end{array}\right|=(1- \lambda)(3- \lambda)- 4= \lambda^2- 4\lambda- 1= 0$. $\lambda= \frac{4\pm\sqrt{16+ 4}}{2}= \frac{2\pm \sqrt{5}}{2}$.

    Now look for the corresponding eigenvectors. If (x, y) is an eigenvector corresponding to eigenvalue $\frac{2+ \sqrt{5}}{2}$ then we have $\begin{bmatrix}2 & -2 \\ -2 & 3\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}= \begin{bmatrix}2x- 2y \\ -2x+ 3y\end{bmatrix}= \begin{bmatrix}\frac{2+ \sqrt{5}}{2}x \\ \frac{2+ \sqrt{5}}{2}y\end{bmatrix}$. We must have $2x- 2y= \frac{2+ \sqrt{5}}{2}x$ and $-2x+ 3y= \frac{2+ \sqrt{5}}{2}y$. You should (modulo any arithmetic errors on my part) find that those two equations are equivalent and you can solve for y in terms of x and get a solution for any value of x.
    Thanks from s_ingram and topsquark
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    Re: what's in a name?

    Thanks for you detailed reply, HallsofIvy.
    I see what you mean.
    I cannot look at an equation and deduce what shape it is (without a graphical tool) because the shape could be a rotation, reflection or translation and its equation therefore not recognisable without analysis.
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    Re: what's in a name?

    Quote Originally Posted by DenisB View Post
    Hey Walagaster: can I borrow your cup?
    Dude!! That's no "cup". It's ein Klein stein.
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    Re: what's in a name?

    Quote Originally Posted by Walagaster View Post
    Dude!! That's no "cup". It's ein Klein stein.
    "Cheap, fast, and reliable. Pick any two." :
    OK, I will; 2 that are full of quality Arizona whiskey!
    Last edited by DenisB; May 15th 2018 at 04:46 PM.
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