Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 20 / b

The curve has $x\ne 0$ and can be written $y=1/x-x$. It is a hyperbola with vertical asymptote $x=0$ and oblique asymptote $y=-x$, crossing the horizontal axis at $(\pm 1,0)$. It is symmetric under $(x,y)\mapsto (-x,-y)$.

On each branch $dy/dx=-1/x^2-1<0$. For $x>0$, the branch descends from $+\infty$ as $x\downarrow 0$, passes through $(1,0)$, and approaches $y=-x$ from above as $x\to +\infty$. The negative branch follows by the central symmetry.

Minimize the squared distance $F=x^2+y^2$, which has the same minimizers as the distance. The Lagrange multiplier equations for the constraint $G=x^2+xy-1=0$ areThe 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 $t=-1\pm \sqrt{2}$. But the constraint says $x^2(1+t)=1$, requiring $1+t>0$ and selecting $t=\sqrt{2}-1$. It follows thatTo prove a global minimum on the whole $x>0$ branch, substitute $y=1/x-x$ into $F$: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.