The shortest distance between the line $x-y=1$ and the curve $x^2=2y$ is

Consider the curve $x^2=2y$

            Differentiate both sides with respect to $x$

$\begin{array}{c}2x=2\frac{dy}{dx}\\ \frac{dy}{dx}=x\end{array}$

slope of the given line x-y=1 is 1 

therefore x=1  

Point on the curve   $P\left(1,\frac{1}{2}\right)$     ( since x=1 $\Rightarrow$ 2y=1 )

Eq of tangent to the curve at P is 

$y-\frac{1}{2}=1(x-1),y=x-\frac{1}{2}\Rightarrow x-y=\frac{1}{2}$

Shortest distance = distance between the parallel lines x-y=1 & x-y=$\frac{1}{2}$is  $=\left|\frac{1-\frac{1}{2}}{\sqrt{1+1}}\right|=\boxed{\frac{1}{2\sqrt{2}}}$