Find the point P on the graph of the function y=\sqrt{x} closest to the point (2,0)

Answer 1

at x, the tangent has slope 1/(2√x), so the normal has slope -2√x
So, the normal line at (k,√k) is

y-√k = -2√k (x-k)
Since (2,0) is on that line,

0-√k = -2√k(2-k)
1/2 = 2-k
k=3/2

The point is (3/2,√(3/2))
==================================
or, just using the distance, we want the distance

d = √((2-x)^2+(0-√x)^2)
= √(x^2-3x+4)
to be a minimum

dd/dx = (2x-3) / 2√(x^2-3x+4)
dd/dx=0 at x = 3/2
The desired point is (3/2,√(3/2))