The shortest distance between the point $(\frac{3}{2},0)$ and the curve $y=\sqrt{x},(x>0)$ is

The correct option is B $\frac{\sqrt{5}}{2}$
Let points $(\frac{3}{2},0),(t^2,t)t>0$
Distance $=\sqrt{t^2+{(t^2-\frac{3}{2})}^2}$
$=\sqrt{t^2-2t^2+\frac{9}{4}}=\sqrt{(t^2-1{)}^2+\frac{5}{4}}$
So minimum distance is $\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$.