<s>[INST] Find the point on the curve $x^{2} - yz=1$ that is closest to the origin The problem statement is as follows:
Find the point on the curve $x^{2} - yz=1$ that is closest to the origin.
To do this we minimize $D^{2} = x^{2} + y^{2} + z^{2}$ with the above constraint.
Attempted solution:
Isolate $z$ in the constraint:
$$
x^{2} - yz = 1 \implies z = \frac{x^{2}-1}{y}
$$
Substitute into distance formula:
$$
x^{2} + y^{2} + \left(\frac{x^{2}-1}{y}\right)^{2}
$$
Differentiate in terms of x and y and set to zero
$$
\frac{\partial D^{2}}{\partial x} = 2x+4x\left(\frac{x^{2}-1}{y^{2}}\right)=0,\frac{\partial D^{2}}{\partial y} = 2y-2\left(\frac{(x^{2}-1)^{2}}{y^{3}}\right)=0
$$
which has real solutions $x=0, y=+1$, giving the critical point as $x=0, y=1, z=-1$.
The correct answer is $x=\pm 1, y=0, z=0$.
What am I doing wrong?
 [/INST] In the case of a quadratic form compared with the standard (squared) distance $x^2 + y^2 + z^2,$ from Lagrange multipliers: the extremal directions are the eigenvectors of the Hessian matrix of the other form, here $x^2 - yz \; . \; \;$  In this case the eigenvalues are three distinct integers, one may take eigenvectors with all integer coordinates, find them explicitly, see what drops out...
$$
\left(
\begin{array}{rrr}
2&0&0 \\
0&0&-1 \\
0&-1&0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
0&0&1 \\
1&-1&0 \\
1&1&0 \\
\end{array}
\right) =
\left(
\begin{array}{rrr}
0&0&2 \\
-1&-1&0 \\
-1&1&0 \\
\end{array}
\right)
$$
The locus $x^2 -yz=1$  is a hyperboloid of one sheet, the eigenvector $(0,1,1)^T$ is the axis, multiples of it never lie on your surface.
</s>