= Solution
Minimize the squared distance $F=x^2+y^2$, which has the same minimizers as the distance. The <Lagrange multiplier> equations for the constraint $G=x^2+xy-1=0$ are
$$
2x=\lambda(2x+y),\qquad 2y=\lambda x,\qquad x^2+xy=1.
$$
The constraint <gradient> cannot vanish on this branch, since its second component is $x>0$. Eliminate $\lambda=2y/x$ to obtain $x^2=2xy+y^2$. Put $t=y/x$; then $t^2+2t-1=0$, so $t=-1\pm\sqrt2$. But the constraint says $x^2(1+t)=1$, requiring $1+t>0$ and selecting $t=\sqrt2-1$. It follows that
$$
\boxed{(x,y)=\left(2^{-1/4},(\sqrt2-1)2^{-1/4}\right),\qquad
d_{\min}=\sqrt{2(\sqrt2-1)}.}
$$
To prove a <global minimum> on the whole $x>0$ branch, substitute $y=1/x-x$ into $F$:
$$
F=2x^2+x^{-2}-2\geq2\sqrt2-2.
$$
The <arithmetic-geometric mean inequality> proves this bound, with equality only when $2x^2=x^{-2}$, namely $x=2^{-1/4}$. Thus the <stationary point> is the unique closest point on the required branch. Numerically it is approximately $(0.840896,0.348311)$ and its distance is $0.910180$ in coordinate units.
Back to article page
