
> 
> What is the shortest curve $\gamma$ in $\mathbb{R}^2$
>  from the origin $o=(0,0)$ to a rational point $p=(a,b)$
>  that (a) passes through no other rational point, and
>  (b) contains no point a rational distance from both $o$ and $p$?
> 
> 
> 


A rational point is one with rational coordinates.
I am wondering if one can reach $p$ via such a highly
"irrational route."
My guess is that there are curves whose lengths approach
the straight-line distance $|p|$.
Perhaps the curve should be restricted to a specific class:
$C^\infty$, analytic, elliptic, quadratic, circle arc.
(Just avoiding rational points [condition (a)] can be accomplished with a
circle arc of an appropriately chosen radius.)


This is far from my experience, and it may be that all
the variations have trivial, uninteresting answers, in
which case I apologize for the distraction.



Shortest irrational path
The simplest smooth curve that avoids all rational points is probably the parabola


$$y = \frac{b}{a}x + \lambda x(a-x)$$


where $\lambda$ is any irrational number. 


Now, the set of points in $\mathbb{R}^2$ that are a rational distance from both $o$ and $p$ is countable (because any two rationals determine at most two such points); but the set of irrational $\lambda$ is uncountable. Moreover, as $\lambda$ varies, the resulting parabolas are all disjoint (apart from their end-points $o$ and $p$). So there must exist (uncountably many) $\lambda$ for which the above parabola satisfies your conditions. Also, as you suspect, $\lambda$ can be chosen as close to zero as you like.


**Updated** In fact we can take any one-parameter family of curves $\{C\_{\lambda}\}$ from $o$ to $p$ which are disjoint except at $o$ and $p$. The set of all rational points, together with the set of points that are a rational distance from both $o$ and $p$, is countable, so in any open interval $I$ there exists $\lambda \in I$ such that $\{C\_{\lambda}\}$ satisfies OP's conditions (a) and (b).


