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High School student here. I have been trying to derive a formula for an egg shaped curve. There are plenty of examples of egg shaped curves in the internet. Apparently it is a common topic.

I noticed Mr. Itou's approach for said curve as a section made by cutting a Pseudo-sphere by means of inclined plane. (link, bottom of the page). I did not see the advantage in using a pseudo-sphere instead of a simpler surface.

I decided to do something similar, but starting with an hyperbolic funnel (Surface of revolution made by rotating a hyperbola in the ZY plane arround the Z-axis).

I wrote the equation, and after the rotation I obtained this parametric equation:

$$ \begin{cases} & x = t\sin s\\ & y = t\cos s \\ & z = 1/t \end{cases} $$

After eliminating the parameters (using the pitagorean identity) I arrive at the following equation:

$x^{2}y^{2}+z^{2}y^{2}-1=0$

After solving for z, and plot for ${z \in \mathbb{R}:z>0}$ I got:

Function plot

Now I would like to intersect the surface with an inclined plane to obtain my egg shaped curve. Then I would rotate the $P(x,y,z)$ to the XY plane to get a function $f(x)$.

I have no idea how to do this (obtaining the curve equation)

I tried equating both expressions (the surface and the plane) but when I plot it I get a strange curve. I tried to intersect it with the plane $z=3$ but instead of a circle I get something resembling a Gaussian curve reflected on the x-axis.

$x^{2}y^{2}+z^{2}y^{2}-z-1=0$

I have some knowledge in linear algebra and working with planes, lines in $R^{3}$.

I made the intersection by hand (technical drawing is pretty useful :) so I know that my original equations are right.

Any hints on how to get the function of the intersection?

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It is easy to intersect your surface with a plane of equation $z=k$: inserting here your third parametric equations one gets $t=1/k$ and then you have the parametric equations of the intersection curve, which is obviously a circle: $x=(1/k)\sin s$ and $y=(1/k)\cos s$.

The intersection with a more general plane can be done in the same way: if the equation of the plane is $ax+by+cz=k$, insert here the parametric equations, solve for $t$ and insert its expression (which will be now a function of $s$, in general) back into the parametric equations, which now depend only on $s$ and thus describe a curve in space.

Of course to obtain a plane equation of that curve you'll have to rotate the frame, so that normal vector $(a,b,c)$ is rotated along the $z$ axis.

EDIT

I did the above with a generic plane $\sin\alpha\ x+\cos\alpha\ z = k$ and got the following parametric equation in the plane: $$ \begin{cases} & x = {\displaystyle k \cot\alpha \left(\sqrt{k^2-2 \sin 2\alpha \sin s}-k\right)+2 \sin s \over\displaystyle \sqrt{k^2-2 \sin 2\alpha) \sin s}-k} \\ &\\ & y = \frac{1}{2} \csc\alpha \cot s \left(k-\sqrt{k^2-2 \sin2\alpha \sin s}\right) \end{cases} $$ This curve is closed if $k^2\ge2\sin2\alpha$ and it is clearly egg-shaped only for $k$ very near to its limiting value. The following picture, for instance, was obtained with $\alpha=\pi/12$ and $k=1.001$:

enter image description here

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  • $\begingroup$ You are fast! I was still exploring your suggestion when you edited. I was working on MATLAB numerically plotting both the surface and the plane to find the "optimal" values of $k$ and alpha. (optimal meaning subjectively closer to an egg shape). $\endgroup$ – mmtz Dec 8 '15 at 18:01
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The equation of hyperbolic horn should be:

$$ ( x^2+ y^2) z^2 = 1 \tag{1}$$ Accordingly you may please like to change your question.

EDIT 1:

Let the cutting plane be

$$ z = m x + c \tag{2}$$

The space curve on a hyperbolic horn can be described with a single parameter $u$:

$$ x = u ;\, y= \sqrt {\frac { 1}{ (m u +c)^2 } - u^2 };\, z = m u + c ; \,$$

(By iteration we can compute a pseudospherical horn intersection as well numerically ).

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  • $\begingroup$ This sounds like a request for clarification more than an answer. Perhaps a bit of added explanation would help your Readers? $\endgroup$ – hardmath Jan 29 '16 at 3:45
  • $\begingroup$ So only a hint please. $\endgroup$ – Narasimham Jan 29 '16 at 6:25

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