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

The correct option is A $\frac{\sqrt{5}}{2}$
Any point on the curve $y=\sqrt{x}$ will be of the form $(t^2,t),t>0$
Using distance formula
$d=\sqrt{(t-0{)}^2+{(t^2-\frac{3}{2})}^2}$
$d=\sqrt{t^2+t^4-3t^2+\frac{9}{4}}$
$d=\sqrt{(t^2-1{)}^2+\frac{5}{4}}$
$d$ will be minimum when $t^2=1$
$d_{\text{min}}=\frac{\sqrt{5}}{2}$