= Solution
The curve is a <hyperbola> with two branches, vertices $(0,1)$ and $(0,-1)$, and asymptotes $y=\pm x$. It is symmetric under <reflection> in either coordinate axis. The requested sketch marks both vertices:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ia/paper-3-hyperbola.png]
{title=Two branches of the hyperbola, with its asymptotes and the points of minimum radius of curvature}
Parametrize the branches by $x=\sinh t$, $y=\pm\cosh t$, for $t\in\mathbb R$. The squared speed is $\dot x^2+\dot y^2=\cosh^2t+\sinh^2t=\cosh2t$, and $|\dot x\ddot y-\ddot x\dot y|=|\cosh^2t-\sinh^2t|=1$. The reciprocal of the <curvature of a plane curve> is therefore
$$
\boxed{\mathcal R(t)=(\cosh2t)^{3/2}=(1+2x^2)^{3/2}}.
$$
Since $\cosh2t\ge1$, with equality only at $t=0$, \b[the least radius is $1$, attained at $(0,1)$ and $(0,-1)$].
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