question_answer

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

A) \[\frac{\sqrt{5}}{2}\]                             

B)  \[\frac{\sqrt{3}}{2}\]

C) \[\frac{3}{2}\]              

D)        \[\frac{5}{4}\]

Correct Answer: A

Solution :

Let P be the point nearest to \[\left( \frac{3}{2},0 \right),\]then normal at

P will pass through \[\left( \frac{3}{2},0 \right).\]

Let Co-ordinates of P be \[s\left( \frac{{{t}^{2}}}{4},\frac{t}{2} \right)\]

Hence, equation of normal is  \[y+tx=\frac{t}{2}+\frac{{{t}^{2}}}{4}\]

The line passes through \[\left( \frac{3}{2},0 \right)\]

\[\frac{3t}{2}=\frac{t}{2}+\frac{{{t}^{3}}}{4}\]\[\Rightarrow \]t = 2 \[(-2,0\] are rejected)

Hence, nearest point is (1 ,1)

Distance \[\sqrt{{{\left( \frac{3}{2}-1 \right)}^{3}}+{{(1-0)}^{2}}}=\frac{\sqrt{5}}{2}.\]