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General Background

Suppose I have one side of a two-sheet hyperboloid as a general three dimensional shape, where the symmetry axis is along, say, my x-axis ($\hat{\mathbf{x}}$) in my chosen coordinate basis. If I look at the intersection line in either the XY- or YZ-planes, the line can be given (in polar coordinates) by: $$ r + x_{o} = \frac{ L }{ 1 + e \cos{\theta} } \tag{1} $$ where $r$ is the distance from the focus to the path, $x_{o}$ is the displacement of the focus along $\hat{\mathbf{x}}$ relative to the origin, $L = a \ \left( 1 - e^{2} \right) = b^{2}/a$ is the semi-latus rectum, $e$ is the eccentricity, $\theta$ is the polar angle, $a$ is the semi-major axis and $b$ the semi-minor axis. Since $e > 1$, the line demarking the intersection of the hyperboloid with the plane of projection is clearly a hyperbola.

Assume that so long as the plane of projection contains the symmetry axis, Equation 1 will always describe the intersection of the hyperboloid with the plane of projection (i.e., just assuming azimuthal symmetry). In other words, Equation 1 will generate a two dimensional surface if rotated about the $\hat{\mathbf{x}}$-axis.

Problem

The Equation 1 works if the plane of projection contains the symmetry axis. However, suppose I want to find a general equation, of similar form to Equation 1, that defines the intersection line for arbitrary offsets of the plane of projection. To keep things simple, assume that the offset is only along either $\hat{\mathbf{y}}$ or $\hat{\mathbf{z}}$, not both simultaneously (i.e., not looking for a completely general form allowing for arbitrary offsets in any direction, just one direction at a time).

Questions

  1. Is there a closed form describing the intersection line for arbitrary offsets along only the $\hat{\mathbf{y}}$- or $\hat{\mathbf{z}}$-directions?
    • If no to the first question, then is there an abbreviated form that would simplify any numerical analysis?
    • If yes to the first question, what is the result and how is it derived?

Side Note

My purpose for looking into this is to create a simple routine that will plot the location of a bow shock given an arbitrary plane of projection offset along either the $\hat{\mathbf{y}}$- or $\hat{\mathbf{z}}$-directions.

Thus, as the magnitude of the offset increases from zero I expect the magnitude of $r + x_{o}$ to change (and perhaps even the shape of the curve?). I would guess that one could construct a general form where $r = r(y_{o},z_{o})$ and $x_{o} = x_{o}(y_{o},z_{o})$ in that plane, but I am not sure how.

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  • $\begingroup$ If I understand correctly, you have a hyperboloid generated by the rotation of a hyperbola around one of its axes of symmetry, and want to find its intersection with a plane parallel to that axis. If so, the answer is quite simple: the intersection is always a hyperbola. $\endgroup$ – Aretino Jan 21 '16 at 21:25
  • $\begingroup$ @Aretino - Yes, I know it's a hyperbola, but I am curious how the parameters in Equation 1 change or depend upon the y or z offset. $\endgroup$ – honeste_vivere Jan 21 '16 at 22:57
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In principle it is quite simple to find what you need, one only needs to discuss two different cases: whether the hyperboloid is of one or two sheets. I'll treat the one sheet case, the other is completely analogous.

The equation of a one-sheet hyperboloid, symmetric around the $x$ axis is: $$ {y^2+z^2\over a^2}-{x^2\over b^2}=1. $$ Intersecting that with the $(x,y)$ plane (equation $z=0$) gives the equation of a hyperbola of equation $y^2/a^2-x^2/b^2=1$, with semi-major axis $a$ and semi-minor axis $b$.

Intersecting instead with the a plane parallel to the $(x,y)$ plane and at a distance $d$ from it (equation $z=d$) gives the equation of a hyperbola of equation $y^2/a^2-x^2/b^2=1-d^2/a^2$, that is: $$ {y^2\over\beta a^2}-{x^2\over\beta b^2}=1, $$ where $\beta=1-d^2/a^2$.

If $\beta>0$ (that is $d<a$) this is the equation of a hyperbola intersecting the $y$ axis, with semi-major axis $\sqrt\beta a$ and semi-minor axis $\sqrt\beta b$.

If $\beta<0$ ($d>a$) this is the equation of a hyperbola intersecting the $x$ axis, with semi-major axis $\sqrt{-\beta} b$ and semi-minor axis $\sqrt{-\beta} a$.

If $\beta=0$ ($d=a$) the curve degenerates into a pair of straight lines of equations $y=\pm (a/b)x$, which are also the common asymptotes of all the hyperbolas obtained for $\beta\ne0$.

Notice that the eccentricity of these hyperbolas is fixed: $$ e={\sqrt{|\beta| a^2+|\beta| b^2}\over \sqrt{|\beta|} a}={\sqrt{a^2+ b^2}\over a}, $$ while the semi-latus rectum gets multiplied by $\sqrt{|\beta|}$: $$ L={(\sqrt{|\beta|} b)^2\over\sqrt{|\beta|} a}=\sqrt{|\beta|} {b^2\over a} \ \ \hbox{for $\beta>0$, while}\ \ L=\sqrt{|\beta|} {a^2\over b} \ \ \hbox{for $\beta<0$.} $$

Foci lie on the $y$ axis if $\beta>0$, and on the $x$ axis if $\beta<0$, at a distance $\sqrt{|\beta|}\sqrt{a^2+b^2}$ from the origin.

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  • $\begingroup$ Thanks for the answer. I guess I should have been more clear (will correct shortly). In analogy to a bow shock, I would be looking at one side of a two-sheet hyperboloid. Though it looks like the result will be similar. Is there a closed form solution when one converts this to polar coordinates? $\endgroup$ – honeste_vivere Jan 22 '16 at 13:55
  • $\begingroup$ Notice first of all that I've just corrected a typo in my answer, as I forgot that $\sqrt\beta$ should appear in the expressions for the semi-axes. The polar equation can be recovered expressing $L$ and $e$ in terms of $a$ and $b$, as already indicated in your question. As for $x_0$, I must confess I haven't understood what it represents. $\endgroup$ – Aretino Jan 22 '16 at 14:18
  • $\begingroup$ Oh sorry, it's just an offset of the origin from the focal point along the axis of symmetry. It is useful in my particular example because the bow shock I am examining is created by the terrestrial magnetosphere, which shifts the focal point away from Earth centered. Since many coordinate bases are Earth centered, the offset is necessary for modeling the bow shock... $\endgroup$ – honeste_vivere Jan 22 '16 at 14:26
  • $\begingroup$ See my edit for the expressions of $e$ and $L$. $\endgroup$ – Aretino Jan 22 '16 at 14:39
  • $\begingroup$ Excellent! And since the $x_{o}$ is a constant, I am thinking that should not depend upon the offsets... Hmm... $\endgroup$ – honeste_vivere Jan 22 '16 at 14:43
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Suppose I have one side of a two-sheet hyperboloid as a general three dimensional shape, where the symmetry axis is along, say, my x-axis ($\hat{\mathbf{x}}$) in my chosen coordinate basis.

Ok, so below we see the hyperboloid and plane $$ -\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = -1 \\ z = z_0 $$ for $a = b = c = 1$, $z_0 = 2$ and their intersection curve.

hyperboloid 2 sheet

This means in the plane $z=z_0$ we have: $$ -\frac{x^2}{a^2} + \frac{y^2}{b^2} = -\left( 1 + \frac{z_0^2}{c^2} \right) \iff \\ y = \pm b \sqrt{\frac{x^2}{a^2} - \left(1 + \frac{z_0^2}{c^2}\right)} $$

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  • $\begingroup$ Yes, I can do that in my head but it does not tell me the form of the intersection curve as a function of the offsets. $\endgroup$ – honeste_vivere Jan 22 '16 at 14:33
  • $\begingroup$ Okay, the edit helps... $\endgroup$ – honeste_vivere Jan 22 '16 at 14:44

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