Question

Find the point on the function
y= 1/ x, x>0 that is closest to the point (2,2) and then state the minimum distance.

Answer 1

Let P(x,y) be that point.

A sneaky and easy way to do this question is to make use of the fact that at P, the slope of the line from P to (2,2) must be perpendicular to the tangent at P to the curve.
Slope of line to (2,2) is (y-2)/(x-2)

dy/dx = -1/x^2, so at (x,y)
the slope of the tangent is -1/x^2
for perpendicular lines, slopes are opposite reciprocals of each other.
so
(y-2)/x-2) = +x^2
y-2 = x^2(x-2)
but y = 1/x
1/x = x^3 - 2x^2 + 2
1 = x^4 - 2x^3 + 2x
x^4 - 2x^3 + 2x - 1 = 0

Just looking at this, we can see that x=1 works
If x=1 then y = 1/1 = 1
so P is (1,1)
(looking at my sketch, this makes perfect sense)

so the shortest distance is √(2-1)^2 + (2-1)^2 = √2