JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Application of Derivatives

The line $y=x$ comes closest to the parabola $y^2=x-2$ where the parabola's tangent runs parallel to the line. Differentiating $y^2=x-2$ gives $2y\frac{dy}{dx}=1$, so the slope is $\frac{1}{2y}$; setting it equal to $1$ gives $y=\frac{1}{2}$ and $x=y^2+2=\frac{9}{4}$. The nearest point is $\left(\frac{9}{4},\frac{1}{2}\right)$, whose distance to the line $x-y=0$ is $\frac{\left|\frac{9}{4}-\frac{1}{2}\right|}{\sqrt{2}}=\frac{7/4}{\sqrt{2}}=\frac{7}{4\sqrt{2}}$.