椭球面与三正交切面问题

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证明由椭球$ax^2+by^2+cz^2=1$的三个相互垂直的切平面的交点的轨迹为与椭球同心的球体。
socratic.org/questions/show-that-the-path-tra … ual-perpendic#396037
Calling $E→f(x,y,z)=ax^2+by^2+cz^2-1=0$

If $p_i = (x_i,y_i,z_i)\in E$ then

$ax_ix+by_iy+cz_iz=1$ is a plane tangent to $E$ because has a common point and $\vec n_i = (ax_i,by_i,cz_i)$ is normal to $E$

Let $Π→\alpha x+\beta y+\gamma z=\delta$ be a general plane tangent to $E$ then

$x_i=\frac{\alpha}{a\delta},y_i=\frac{\beta}{b\delta},z_i=\frac\gamma{c\delta}$

but

$ax_i^2+by_i^2+cz_i^2=1$ so

$\frac{\alpha^2}a+\frac{\beta^2}b+\frac{\gamma^2}c=\delta^2$ and the generic tangent plane equation is

$\alpha x+\beta y + \gamma z=\pm\sqrt{\frac{\alpha^2}a+\frac{\beta^2}b+\frac{\gamma^2}c}$

Now given three orthogonal planes

$Π_i→\alpha_i x+\beta_i y+\gamma_i z=\delta_i$

and calling $\vec v_i = (\alpha_i ,\beta_i ,\gamma_i)$ and making

$V=(\vec v_1,\vec v_2,\vec v_3)$ we can choose

$V \cdot V^T = I_3$

and as a consequence

$V^T\cdot V = I_3$

then we have also

$\sum_i\alpha_i^2 =1,\sum_i\beta_i^2 =1,\sum_i\gamma_i^2 =1,\sum_i\alpha_i\beta_i=0,\sum_i\alpha_i\gamma_i=0,\sum_i\beta_i\gamma_i =0$

Now adding $\sum_i(\alpha_i x+\beta_iy+\gamma_iz)^2$ we have

$x^2\sum_i \alpha_i^2+y^2\sum_i \beta_i^2+z^2\sum_i \gamma_i^2+2(xy \sum(\alpha_i \beta_i)+zx\sum(\alpha_i \gamma_i)+yz\sum(\beta_i \gamma_i))=\sum_i \delta_i^2$

and finally

$x^2+y^2+z^2=\sum_i\delta_i^2$

but $\sum_i \delta_i^2={\sum_i\alpha_i^2\over a}+{\sum_i\beta_i^2\over b}+{\sum_i\gamma_i^2\over c}=\frac1a+\frac1b+\frac1c$

so

$x^2+y^2+z^2=\frac1a+\frac1b+\frac1c$

which is the path traced by the point of intersection of three mutual perpendicular tangent planes to the ellipsoid.

Attached a plot for the ellipsoid

$x^2+2y^2+3z^2=1$

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