Question

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

• A. $\frac{\sqrt{5}}{2}$
• B. $\frac{5}{4}$
• C. $\frac{3}{2}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (A)

$\frac{\sqrt{5}}{2}$

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 $\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$.