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

Solution :
Let $$\left( {{a^2},\,a} \right)$$   be the point of shortest distance on $$x = {y^2}.$$   Then distance between $$\left( {{a^2},\,a} \right)$$   and line $$x-y+1=0$$   is given by
$$D = \frac{{{a^2} - a + 1}}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\left[ {{{\left( {a - \frac{1}{2}} \right)}^2} + \frac{3}{4}} \right]$$
It is minimum when $$a = \frac{1}{2}$$   and $${D_{\min .}} = \frac{3}{{4\sqrt 2 }} = \frac{{3\sqrt 2 }}{8}$$