JEE Main · 2009 · 12 Apr 2009 · Offline · Application of Derivatives

The shortest distance from the line to the parabola occurs where the tangent to $x=y^2$ runs parallel to the line $y-x=1$, whose slope is 1. Differentiating $x=y^2$ gives $\frac{dy}{dx}=\frac{1}{2y}$, and setting this equal to 1 gives $y=\frac{1}{2}$ and then $x=\frac{1}{4}$, so the nearest point on the curve is $\left(\frac{1}{4},\frac{1}{2}\right)$. The distance from this point to the line $x-y+1=0$ is $\frac{|\frac{1}{4}-\frac{1}{2}+1|}{\sqrt{1^2+1^2}}=\frac{3/4}{\sqrt{2}}=\frac{3\sqrt{2}}{8}$.