Question

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

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

Answer 1

Answer: (A)

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