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 $S$ ?
What is the formula of the following?
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0It is the check mark to the left of each answer – 2012-04-24
1 Answers
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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0It seems to me okay if we consider the case ellipse in $xy$-plane. – 2012-04-24