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Motivation: Develop some intuition on the solution of the intersecting points of two ellipsoids.

In two dimensions, two arbitrary ellipses can intersect in (at most) four points (excepting the pathological perfect overlap). If you move one ellipse away you can get two of the points to reduce to a single point. Moving further, you can remove this point leaving only two total points of intersection. Finally, you can reduce these two points the same way until you have zero points of intersection. By writing down the equations, it seems that this could be understood as the zeros of a quartic polynomial, where the number of real solutions correspond to the number of points of intersection $f(x)=0$.

In three dimensions, there are now parametrized curves of intersection rather then points. Is it correct to consider the solution of two ellipsoids as closed curves (or points) in a two dimensional plane of a two dimensional polynomial $f(x,y)=0$?

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"If you move away you can get two of the points to reduce to single point" I am taking it that by this you mean two ellipses can intersect at 3 points. But it is'nt possible since a quartic polynomial with real coefficients cannot have exactly 3 real roots(Geometric intuition is not always accurate!). – Dinesh Nov 2 '11 at 22:01
@Dinesh What about $f(x)=x^2 (x-1) (x+1)$ ? Clearly there are three real unique roots, with a degeneracy on the root at zero. I think I wasn't clear as I intended the number of unique real roots. My geometric intuition still tells me that the roots become degenerate as the two points of intersection become a single point. – Hooked Nov 3 '11 at 0:46
"Is it correct to consider the solution of two ellipsoids as closed curves (or points) in a two dimensional plane" - you do know you can set things up such that the intersection of two ellipsoids is a pair of noncoplanar curves, yes? – J. M. Nov 3 '11 at 1:31
@J.M. I think that was what I was getting out, but no, I haven't seen that before. Mind posting it as an answer to give more depth? – Hooked Nov 3 '11 at 2:30
I'll just leave this here... – J. M. Nov 3 '11 at 2:37
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Let the equations of the ellipsoids be $f(x,y,z)=0$ and $g(x,y,z)=0$ ,here $f$ and $g$ are $2$nd degree polynomials in 3 variables.( I avoid writing them explicitly).

By eliminating one variable, (say $z$) we get a 2nd degree polynomial $h(x,y)=0$ (You may want to consider some coordinate transformations to make this easy) which can be a curve or a point(s) as desired.

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Write $f(x,y,z)=f_0 z^2 + f_1 z + f_2 = 0$ and $g(x,y,z) = g_0 z^2 + g_1 z + g_2 = 0$, where each $f_i$ and $g_i$ is a degree-$i$ polynomial in $x$ and $y$. Then $g_0 f - f_0 g = z (g_0 f_1 - g_1 f_0 ) + (g_0 f_2 - f_0 g_2) = z p_1 + p_2$, where $p_i$ is a degree-$i$ polynomial in $x$ and $y$. Now, $z= -p_2/p_1$, so that (plugging into the $f=0$ equation and clearing fractions) we have $f_0 p_2^2 - f_1 p_1 p_2 + f_2 p_1^2 = 0$. Thus, eliminating $z$ gives a 4th-degree polynomial in $x$ and $y$, not a 2nd-degree one. – Blue Dec 3 '11 at 1:13

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