The correct option is A (1,1)
Given, parabola is y=x2 ....(i)
and straight line is y=2x−4 ....(ii)
From equations (i) and (ii), we get
x2−2x−4=0
Let f(x)=x2−2x−4
Thus f′(x)=2x−2
For least distance, put f′(x)=0
⇒2x−2=0
⇒x=1
From equation (i), we have y=1
Hence, the point least distant from the line is (1,1).