Show that greatest value of shortest distance between axis of $x$ and a normal to ellipsoid is $$b-c$$
or in other words show that the maximum distance of all the normals to the ellipsoid is $b-c$
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
My attempt :
I have tried to find shortest distance between lines $ \vec{a_{1}}+\vec{b_{1}}t $
and $ \vec{a_{2}}+\vec{b_{2}}t $
using
$$ (a1-a2). \frac{\vec{a_{1}}*\vec{a_{2}}}{|\vec{a_{1}}*\vec{a_{2}}|}$$
Assuming that the point at which I am drawing the normal is $(\alpha,\beta,\gamma)$ it comes out to be
$$\gamma\beta \frac{b^2-c^2}{b^2c^2*\sqrt{\frac{\gamma^2}{c^4}+
\frac{\beta^2}{b^4}}}$$
I am not able to go further . Any help/hint would be appreciated
Best Answer
Observe that the same values of $x,y,z$ which maximize $$ d(x,y,z)=\frac{|b^2-c^2|}{\sqrt {\dfrac {b^4}{y^2}+\dfrac {c^4}{z^2}}}\tag1 $$subject to the constraint $$ \dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}+\dfrac {z^2}{c^2}=1,\tag2 $$ will minimize the value of $$ \frac {(b^2-c^2)^2}{d^2}=\dfrac {b^4}{y^2}+\dfrac {c^4}{z^2}$$ subject to the same constraint.
The equation $(2)$ together with the other three Lagrange equations for the latter problem: $$\begin {align} \lambda\dfrac {2x}{a^2}&=0;\\ -\dfrac {2b^4}{y^3}+\lambda\dfrac {2y}{b^2}&=0;\\ -\dfrac {2c^4}{z^3}+\lambda\dfrac {2z}{c^2}&=0,\\ \end {align} $$ can be readily solved with the result: $$ x=0,\quad y^2=\dfrac{b^3}{b+c},\quad z^2=\dfrac{c^3}{b+c}, \quad\lambda=\dfrac{b^4}{y^2}+\dfrac{c^4}{z^2}=(b+c)^2. $$ Substituting the last value into $(1)$ one finally obtains: $$d_\text {max}=|b-c|. $$