Given, parabola is y=x2 ... (i)
straight line is y=2x−4 ... (ii) From Eqs. (i) and (ii), we get x2−2x−4=0
Let f(x)=x2−2x−4 ∴f′(x)=2x−2
For least distance, put f′(x)=0 ⇒2x−2=0 ⇒x=1 From Eq. (i), y=1
Hence, the point least distant from the line is (1,1).