# Calculus - 5

The point on the curve $y = \sqrt x$ which is closest to the point $(4,0)$ is

(A) $(4,2)$

(B) $\left ( \frac{7}{2} , \sqrt{\frac{7}{2}} \right)$

(C) $\left ( \frac{1}{2} , {\frac{1}{ \sqrt2}} \right)$

(D) $\left ( \frac{1}{2} , {\frac{3}{ \sqrt2}} \right)$

reshown May 31

from the above options only A,B,C points are the part of curve , now find distance from (4,0) to these options A, B, C . hence correct answer is B
answered May 31 by (780 points)

Distance of (4,0) from any point (x,y) is $\sqrt{(x-4)^2 + (y-0)^2}$ . Here $y=\sqrt{x}$ so

$f(x)=(x-4)^2 + x$  (removing sqrt)

so $f'(x)=2x - 7$

putting $f'(x)= 0$ we get $x = 7 /2$

hence $y=\sqrt{7/2}$

So the point becomes $(7/2,\sqrt{7/2})$

Hence the correct answer is B

answered May 31 by (350 points) 1 flag
edited Jun 3 by
why we removed sqrt at line 2