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Take the ellipsoid for example $$(x^2/a^2)+(y^2/b^2)+(z^2/c^2)=1$$ in the x-y plane you have an ellipse described by $$(x^2/a^2)+(y^2/b^2)=1$$ (suppose z=constant) in the y-z plane you have an ellipse described by $$(y^2/b^2)+(z^2/c^2)=1$$ (x=constant) in the x-z plane you have an ellipse described by $$(z^2/c^2)+(x^2/a^2)=1$$ (y=constant)

I understand how the equation describing a 2d ellipse was derived and I'm pretty sure it can apply to an ellipse in the y-z or x-z plane but I don't understand how the equation describing the 3d ellipse was derived.

I thought about adding all three equations in the three planes but ended up with $$(2x^2/a^2)+(2y^2/b^2)+(2z^2/c^2)=3$$ Dividing by 3 $$(2/3)((x^2/a^2)+(y^2/b^2)+(z^2/c^2))=1$$ $$(2/3)(x^2/a^2)+(2/3)(y^2/b^2)+(2/3)(z^2/c^2)=1$$ Assuming my process is correct would the 2/3 be considered as part of; a on the first term $$(x^2/a^2)$$, b on the second term $$(y^2/b^2)$$ and c on the third term$$(z^2/c^2)$$? resulting in $$x^2/a^2+y^2/b^2+z^2/c^2=1?$$

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Please use LaTeX and change the x^2 to $x^2$ by adding "$" in front of and behind "x^2". –  ᴊ ᴀ s ᴏ ɴ Jul 27 '12 at 5:29

2 Answers 2

I understand the question as follows: can we determine a nondegenerate quadric surface from its traces in the three coordinate planes?

In general, we cannot since all three traces may be empty - for example, take the sphere with radius $1$ centered at $(2,2,2)$. For a less trivial example, consider two paraboloids: circular $z=x^2+y^2$ and elliptic $z=x^2+y^2+xy$. They are different surfaces but we cannot tell them apart by their traces in coordinate planes. There is a similar example with one-sheeted hyperboloids: just replace $z$ with $z^2-1$ on the left.

There is a pattern in the above counterexamples: at least one of the traces was degenerate. Let's assume that all three traces are nondegenerate quadrics - that is, circles/ellipses, parabolas or hyperbolas. Can we recover the surface in this case? The answer is yes and I think I can prove it.

Case 1: the surface does not pass through the origin (which we see from its coordinate traces). Then its equation (which we don't know yet) can be brought into the form $p(x,y,z)=1$ where $p$ has no constant term. We can write the equations of traces as $f(x,y)=1$, $g(x,z)=1$ and $h(y,z)=1$. Now, $p-f$ vanishes when $z=0$, which means it only involves the terms divisible by $z$: namely, $xz$, $z^2$, and $yz$. Of these, the first two are found in $g$ and the last one in $h$. Notice that we are not adding $f,g,h$ together - we merge them by taking the union of all monomials they contain. Something very similar happens in calculus when students are asked to recover a function from its partial derivatives.

Case 2: the surface passes through the origin. Its equation is now $p(x,y,z)=0$ where $p$ still does not have constant term. The trace equations are now $f=0$, $g=0$, and $h=0$ which we can normalize so that at the origin $f_x=g_x$, $f_y=h_y$, and $g_z=h_z$. At most one of these equations may be $0=0$; thus we have enough information to determine the ratios of coefficients with which $f,g,h$ appear in $p$. (This is where nondegeneracy of traces is needed.) Now we merge $f,g,h$ into $p$ as in Case 1.

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Just to clarify why "$p−f$ vanishes when $z=0$" in Case 1. Restricting this polynomial to $z=0$, we find that its zero set contains $(0,0)$ in addition to the $xy$-trace (which does not pass through $(0,0)$). This implies $p-f\equiv 0$ on the plane. –  user31373 Aug 1 '12 at 17:16

(I can't comment since I'm unregistered; feel free to move this into a comment if deemed necessary.)

I don't understand what you mean by "derivation" here, and unlike in the conic section case, there isn't a geometric description for the quadrics that is as simple as "the locus of points whose distance from a fixed point and a fixed line is in a certain ratio $\varepsilon$".

That being said, for the ellipsoid, I like to think of it as an affine deformation of the usual sphere; that is, if you consider three mutually orthogonal axes intersecting at the center of the sphere, and then proceed to "stretch" the sphere along those directions, you will then be obtaining an ellipsoid.

For oblate and prolate spheroids, on the other hand, a geometric view is easier: an oblate spheroid is the surface you get if you rotate an ellipse about its minor axis (resulting in a "squashed" sphere), while a prolate spheroid is the surface of revolution that results when you rotate your ellipse about its major axis (think of something like the ball they use in rugby).

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