Find the smallest distance between point and ellipsoid

Question

Find the smallest distance between the points on the ellipsoid $x^2+2y^2+z^2=16$ and the point $(0,0,1)$.

Answer:

Let $(x,y,z)$ be the closest point of the ellipsoid. The distance is $$\sqrt{x^2+y^2+(z-1)^2}$$ Let $f(x,y,z)=x^2+y^2+(z-1)^2$. Construct the Lagrangian: $$L(x,y,z)=x^2+y^2+(z-1)^2+\lambda (x^2+2y^2+z^2-16)$$ For smallest distance, $$L_x=0 \Rightarrow x(\lambda+1)=0\\ L_y=0 \Rightarrow y(2 \lambda+1)=0\\ L_z=0 \Rightarrow z(\lambda+1)=1$$ The 3rd equation implies that $\lambda\ne-1$. Thus we have $x=0$. Now 2nd and 3rd equation solution gives $y=0$ or $\lambda=-\frac12$ and $z=\frac1{1+\lambda}$.

But I can not find the final solution .

Help me out

Answer 1

Alternative way:

We have found that $x=0$, and we need $|z| \le 4$,

\begin{align}x^2+y^2+(z-1)^2&=\frac{16-z^2}2+(z-1)^2 \\ &=\frac{16+z^2-4z+2}{2}\\ &=\frac{(z-2)^2-4+16+2}{2}\\ &=\frac{(z-2)^2}{2}+7\end{align}

Note that $|2| \le 4$, and hence the smallest distance is $\sqrt{7}$.

Answer 2

If $\lambda=-\frac12$, $z=2$ and from the original constraint $x^2+2y^2+z^2=16$ (where $x=0$ as shown prior) we get $y=\pm\sqrt 6$. In this case, the distance is $d=\sqrt{0^2+\sqrt6^2+(2-1)^2}=\sqrt7$.

If $y=0$, from the original constraint we have $z=4\lor z=-4$, which give $d=\sqrt{0^2+0^2+(4-1)^2}=3$ and $d=\sqrt{0^2+0^2+(-4-1)^2}=5$ respectively.

$\sqrt7$ is thus the shortest distance from $(0,0,1)$ to the ellipsoid, the closest points being $(0,\pm\sqrt6,2)$.