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.
 A: A point lying on the first parabola has coordinates $(1+u^2,u)$ while a point lying on the second parabola has coordinates $(v,1+v^2)$, hence the squared distance between them is given by:
$$ d(u,v) = (1+u^2-v)^2 + (1+v^2-u)^2 $$
and the stationary points for such a function are given by the solutions of:
$$ \frac{\partial d}{\partial u}= 4u(1+u^2-v)-2(1+v^2-u)=0,$$
$$ \frac{\partial d}{\partial v}= 4v(1+v^2-u)-2(1+u^2-v)=0,$$
hence they fulfill, by taking the difference between the two equations:
$$ u(3+u+2u^2)=v(3+v+2v^2)\tag{1} $$
but since $\frac{d}{dt}\left(t(3+t+2t^2)\right) = 3+2t+6t^2$ has a negative discriminant we have that $t\to t(3+t+2t^2)$ is an injective function, hence $(1)$ implies $u=v$ and we have:
$$ (4u-2)(1-u+u^2) = 0 \tag{2}$$
from which $u=v=\frac{1}{2}$ and:
$$  \min_{u,v} {d(u,v)}^2 = 2\cdot\left(1+\frac{1}{4}-\frac{1}{2}\right)^2 = \frac{9}{8}.\tag{3}$$
A: 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.
A: 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 $\sqrt2\,\left|\frac54-\frac12\right|=\frac34\sqrt2$.
A: Hope a bit of calculus is allowed. Primes are with respect to x. Notice that they are inverse functions of each other, you can swap $ x,y$ to get to the second parabola. They are mirror images with respect to line $ x=y$. Required point should have this slope $y^{'} =1 $ for its tangent at point of tangency at ends of common normal.
Take the parabola with its symmetry axis coinciding with axis.
Differentiating  $ 2 y y'= 1 , 2 y = 1, $ and the $x,y$ coordinates are 
$$ (\dfrac54,\dfrac12)$$ 
and the other point of tangency is again swapped to 
$$ (\dfrac12,\dfrac54); $$
Now use distance formula between the tangent points to get
$ d =  \dfrac{3 \sqrt{2}}{4}. $

