TITLE: Minimum distance between two parabolas
QUESTION [15 upvotes]: The shortest distance between the parabolas $y^2=x-1$ and $x^2=y-1$ is.
Attempt: The shortest distance is along the common normal of the two curves.

REPLY [3 votes]: The two parabolas are clearly mirror images by the main diagonal ($x=y$), which diagonal they don't intersect (as $x^2-x+1>0$ for all real $x$). The shortest path then must be perpendicular to that diagonal, and the tangent to the parabolas parallel to the diagonal. The second parabola is the graph of the function $x\mapsto x^2+1$, and no complicated calculus is needed to see that the only point where its tangent is parallel to the diagonal is $(x,y)=(\frac12,\frac54)$. The distance to its mirror image $(\frac54,\frac12)$ is $\sqrt2\,\left|\frac54-\frac12\right|=\frac34\sqrt2$.