<s>[INST] Coordinates of the point of an ellipsoid closest to the origin. Given the ellipse $\frac{(x-r)^2}{a^2}  +  \frac{(y-s)^2}{b^2} =1$ 
$(a^2 > b^2)$. Find the coordinates of the point of the ellipsoid closest to the origin in terms of a,b,r and s.
I tried three different method so far but all lead to highly complicated and lengthy computations. Anyone has a shortcut? 
Thanks
 [/INST] The reason this gets so messy is that the solution depends on solving a quartic equation whose coeffiicents are functions of $a,b,r,s$.
Start by writing the ellipse in parametric form, as 
$$
x= a\cos t + r \\
y= b \sin t + s
$$ 
Then the derivative of the square of the distance is
$$
\cos t \left( 2bs + 2(b^2-a^2) \sin t -2ar \tan t\right)
$$
and unless the solution is at the special point $(x=r,y=s\pm b)$ we can divide by $\cos t$.  Letting $u=\sin t$ and writing $\tan t = \frac{u}{\sqrt{1-u^2}}$
we find that the derivative is zero when
$$
 2bs + 2(b^2-a^2) u = 2ar \frac{u}{\sqrt{1-u^2}}\\
\left(  2bs + 2(b^2-a^2) u\right)^2 = \frac{4a^2r^2u^2}{1-u^2} \\
(b^2-a^2)^2 u^4 + 2(b^2-a^2) bsu^3-\left((b^2-a^2)^2 -a^2r^2 -b^2s^2 \right)u^2 -  2(b^2-a^2)bs u +b^2s^2 = 0
$$
The solution to this quartic can be obtained in closed form but it is spectacularly messy. 
So if you work the problem in any other way (including Lagrange multipliers) that spectacular messiness has to come in to the calculations at some point, becasue it remains at the end of the solution.
</s>