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Let $S$ be the ellipsoid given by the formula $$ \frac{x^2}{a^2}+\frac{y^2}{b^2} +\frac{z^2}{c^2}=1$$ where $a \ge b \ge c > 0$ are fixed constants. What is the formula given by the set consisting of all the intersection points of all triplet pairwise orthogonal tangent planes to the ellipsoid $S$ ?

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Should there be an equality sign somewhere? Or are you just considering a function $(x,y,z)\mapsto \frac{x^{2}}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}$? –  Thomas E. Apr 24 '12 at 13:27
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I guess that $=1$ is missing... –  Siminore Apr 24 '12 at 13:29
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@hkju: When someone answers your question, and you find the answer acceptable, you should click on the checkmark below the vote buttons to say that you "accept" this answer. (And you should probably review your previous questions and do the same thing) –  Najib Idrissi Apr 24 '12 at 13:32
    
I don't currently get what you mean by "the three orthogonal tangent planes". In each point of the ellipsoid you have one unique tangent plane. Or do you mean all triplets of pairwise orthogonal planes that are tangent to the ellipsoid? And of course I second the above three comments. –  Christian Rau Apr 24 '12 at 13:54
    
I am a stranger here. Where is the accept button? –  hkju Apr 24 '12 at 19:52
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up vote 1 down vote accepted

Given a point ${\bf p} = (x,y,z)$, the plane through $\bf p$ with unit normal $\bf u$ is tangent to your ellipsoid iff $(u_1 x + u_2 y + u_3 z)^2 = u_1^2 a^2 + u_2^2 b^2 + u_3^2 c^2$. You want to find all $\bf p$ such that there are three orthonormal solutions to this equation. The set can be parametrized by the orthogonal matrices $U$: if $k_i = \sqrt{u_{i1}^2 a^2 + u_{i2}^2 b^2 + u_{i3}^2 c^2}$ for $i=1,2,3$, the possible $\bf p$ are all the vectors $U^T \pmatrix{k_1\cr k_2\cr k_3\cr}$.

EDIT: Hmm. It looks to me like these $\bf p$ form the sphere centred at the origin with radius $\sqrt{a^2 + b^2 + c^2}$. Well, certainly $${\bf p \cdot p} = (k_1, k_2, k_3) U U^T \pmatrix{k_1\cr k_2\cr k_3\cr} = k_1^2 + k_2^2 + k_3^2 = \sum_{i=1}^3 \left(u_{i1}^2 a_2 + u_{i2}^2 b^2 + u_{i3}^2 c^2 \right) = a^2 + b^2 + c^2 $$

It's not obvious to me why all points on the sphere would be solutions, but it's not surprising that this should be the case.

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It seems to me okay if we consider the case ellipse in $xy$-plane. –  hkju Apr 24 '12 at 23:20
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