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?
 A: Jun has already pointed out your error. I propose a slight different approach where we have no dangerous denominator: we isolate $x^2$ in the constraint, i.e. $x^2=1+yz$ and we consider the squared distance function
$$f(y,z)=(1+yz)+y^2+z^2.$$
Since $f$ is a quadratic polynomial, the derivatives are linear polynomials:
$$\frac{\partial f}{\partial y}=z+2y=0,\qquad \frac{\partial f}{\partial z}=y+2z=0$$
which implies $y=z=0$. Therefore $x^2=1+yz=1$ yields $x=\pm 1$.
A: An alternative approach is using Lagrange multipliers. Construct the Lagrangian auxiliary function
$$\mathcal L=x^{2} + y^{2} + z^{2}+\lambda\left[1-\left(x^{2} - yz\right)\right]$$
and solve the first order conditions
$$\begin{aligned}
\frac{d\mathcal L}{d x}=2x-2x\lambda=0\\
\frac{d\mathcal L}{d y}=2y+z\lambda=0\\
\frac{d\mathcal L}{d z}=2z+y\lambda=0\\
\frac{d\mathcal L}{d \lambda}=1-x^2+yz=0\\
\end{aligned}$$
that, of course, result in $x=\pm 1$, $y=0$ and $z=0$.
A: 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.
