Official solution
Shortest distance must be along common normal
Let a point on curve has x coordinate = h
Then y coordinate :
h2 = 2y
\[ \Rightarrow \] y = \[{{{h^2}} \over 2}\]
So point is = (h, \[{{{h^2}} \over 2}\])
m1 (slope of line x - y = 1) = 1
\[ \Rightarrow \] slope of perpendicular line = -1
Slope of the perpendicular line on the curve x2 = 2y,
\[{m_2} = {{2x} \over 2} = x \Rightarrow {m_2} = h \]
\[ \therefore \] Slope of normal = \[ - {1 \over h}\]
\[ - {1 \over h}\] = -1 \[ \Rightarrow \] h = 1
So point is \[\left( {1,{1 \over 2}} \right)\]
\[D = \left| {{{1 - {1 \over 2} - 1} \over {\sqrt {1 + 1} }}} \right| = {1 \over {2\sqrt 2 }}\]