Minimum distance between two parabolas 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.

let the shortest distance between the parabolas $y = 1 + x^2$ and $x = 1 + y^2$ be given by $AB,$  where $A = (a , 1 + a^2), B = (1 + b^2, b)$ with both $a$ and $b$ positive.  the tangent at $A$ has slope $2a$ and the one at $B$ has slope $\frac 1{2b}$ and the slope of $AB$ is $\frac{1+a^2 - b}{a-1-b^2}$  
we need $$2a = \frac 1{2b} = \frac{1+b^2 - a}{1 + a^2 - b}$$ so that both tangents are parallel and orthogonal to $AB.$ this gives $$b = \frac 1 a, \, 2a = \frac{1 + \frac 1 {a^2} - a}{1 + a^2 - \frac 1 a} = \frac{a^2 + 1 - a^3}{a(a+a^3 -1)} \to 2a^3+2a^5-2a^2=a^2 + 1 - a^3$$ that is $$2a^5+3a^3 - 3a^2 - 1 = 0, a > 0. $$ 
numerically(ti-83 solver), i found  $a = 0.9072$ to be the only solution.
