JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Application of Derivatives

The nearest point on the curve lies where the line from $(\frac{3}{2},0)$ meets the curve at right angles, so I use the normal. Parametrising the curve as $(t^2,t)$, the tangent slope is $\frac{1}{2t}$ and the normal slope $-2t$; the normal through $(t^2,t)$ passing through $(\frac{3}{2},0)$ gives $0-t=-2t(\frac{3}{2}-t^2)$, which tidies to $2t^3-2t=0$, so $t=0,\pm 1$. The meaningful foot is $t=1$, the point $(1,1)$, whose distance from $(\frac{3}{2},0)$ is $\sqrt{(\frac{1}{2}{)}^2+1^2}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$.