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Here is an ellipse, $\mathrm E$, whose center occurs at $x=-1$ and $y=1$ and whose semimajor axis length is $\sqrt {2/5 \,}$. Therefore, the origin is outside of the ellipse. $$\mathrm E = \{ \mathbf x \in \mathbb R^2 : 4x^2 + 3xy + 4y^2 -x +y =0 \}$$

Here it is parameterized in polar coordinates. $$4 \rho ^2 + 3 \rho^2 \cos \phi \sin \phi + 4 \rho^2 + \rho \sin \phi - \rho \cos \phi = 0 \Rightarrow \rho = \frac {\cos \phi - \sin \phi} {4 + 3\cos \phi \sin \phi}$$ Apparently $\rho \in \mathbb R \,\forall \, \phi$. But I would expect that:

$\forall \phi \in \mathcal A \subset \left [0 , 2 \pi \right ], \rho \notin \mathbb R$ ;

$\forall \phi \in \mathcal B \subset \left [0 , 2 \pi \right ]$, the "function" $\rho \left ( \phi \right ) $ should return two real numbers;

$\forall \phi \in \{ m , n \} \subset \left [0 , 2 \pi \right ], \rho \in \mathbb R$.

Why the apparent contradiction?

EricVonB
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    I'm sorry, but I don't understand what you are trying to say. What is $\mathcal{A}$? $\mathcal{B}$? ${m,n}$? Where is the contradiction that you are seeing? – Xander Henderson Oct 19 '18 at 21:31
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    The curve is traced twice as $\phi$ ranges over $[0,2\pi)$. $\rho\lt0$ on half of this interval. – amd Oct 19 '18 at 21:47
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    The origin is on the curve, since the polynomial has a zero constant term. – David K Oct 19 '18 at 22:09
  • Xander Henderson: $ \phi = \arctan \frac {y}{x} $ is the radial coordinate of $\mathbf x \in \mathbb R^2$. The domain $\mathcal A $ is the subset of the interval $\left [0 , 2 \pi \right ] $ for which there exists no $\rho \left ( \phi \right ) $ such that $\left ( \rho , \phi \right ) \in \mathrm E$. – EricVonB Oct 19 '18 at 22:24
  • David K: Which of the following statements is false?:

    The center of $\mathrm E$ is situated at $x=-1$ and $y=1$ ;

    The semimajor axis of $\mathrm E$ has length $\sqrt {2/5 ,}$.

     – EricVonB Oct 19 '18 at 22:28
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    An equivalent equation is $\frac54(x+y)^2-\frac14(x-y)^2-(x-y)=0.$ It's a hyperbola with center $x=-1,y=1,$ semimajor axis $\sqrt2$, the axis lying along the line $y=-x.$ So "length $\sqrt{2/5}$" is false, but also "an ellipse, $E$" is false. – David K Oct 19 '18 at 22:36
  • I am sorry. At first, I typed the wrong coefficients for $x^2$ and $y^2$ in the definition of $\mathrm E$. I have edited the question, such that the polynomial equation that defines $\mathrm E$ contains a positive definite quadratic form; in other words, $\mathrm E$ is an ellipse. – EricVonB Oct 19 '18 at 22:43
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    Now that you've corrected the formula, it is an ellipse, but it still passes through the origin, and the semimajor axis is $\frac15\sqrt2.$ The center is $x=\frac15,y=-\frac15.$ – David K Oct 19 '18 at 22:43
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    Wolfram Alpha can be your friend: https://www.wolframalpha.com/input/?i=4x%5E2+%2B+3xy+%2B+4y%5E2+-x+%2By+%3D0 – David K Oct 19 '18 at 22:46
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    The apparent contradictions lie, as they often do, in your mistaken expectations. You’re not allowing for the fact that $\rho$ can be negative in a polar equation, or that a single line through the origin corresponds to two different values of $\phi$ in the polar equation of the ellipse. – amd Oct 19 '18 at 23:04
  • My understanding of the relation between principal axes lengths and eigenvalues of the quadratic form also appears to be wrong. – EricVonB Oct 19 '18 at 23:11

1 Answers1

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As I mentioned in the comments, you overlooked a couple of sources for the apparent contradictions. First, the curve is traced twice as $\phi$ ranges over $[0,2\pi)$. For half of that range $\rho$ is negative, but it is defined for all values of $\phi$. Second, when working with polar coordinates, you are dealing with rays emanating from the origin. So, when finding the values of $\rho$ that correspond to a given line through the origin, you have to split it into two opposing rays. For each ray, there is only a single value of $\rho$.

The defining equation of $\mathrm E$ is clearly satisfied by $x=y=0$, so the ellipse passes through the origin. Its center is at $(-1/5,1/5)$ and your values for the semiaxis lengths are off, too. I’m not sure what might have gone wrong in your calculation of the center, but based on your comment it appears that you used the reciprocal square roots of the eigenvalues of the quadratic part of the equation. This is correct when the equation has the form $Ax^2+Bxy+Cy^2=1$, but in general you have to apply a scale factor to those eigenvalues: after translation to eliminate the linear terms in the general equation, you have $Ax'^2+Bx'y'+Cy'^2+F'=0$, so you have to divide the quadratic form’s eigenvalues by $-F'$ when computing semiaxis lengths. If you work through this for the general equation, you’ll find that $F'=S/\Delta$, where $S$ is the determinant of the full $3\times3$ matrix of the conic and $\Delta$ is the determinant of the quadratic part—the upper-left $2\times2$ minor of the full matrix and also the product of the quadratic form’s eigenvalues.

amd
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