A line segment has one endpoint, A, on the x-axis and the other endpoint, B, on the y-axis. It passes through the point (1,2). If O is the point (0,0), for what value of angle OAB is the length of the line segment AB a minimum? What is the minimum length?

Answer 1

let A(x,0) , B(0,y) and P(1,2) be the above points
let L = AB

L^2 = x^2 + y^2
but
(1-x)/2 = 1/(2-y)
2 - y - 2x + xy = 2
y(x-1) = 2x
y = 2x/(x-1)

L^2 = x^2 + 4x^2/(x-1)^2
2L dL/dx = 2x - 8x/(x-1)^3 , used quotient rule on last part and simplified

= 0 for a min of L
2x = 8x/(x-1))^3
(x-1)^3 = 4
x-1 = 4^(1/3)
x = 4^(1/3) + 1 = appr 2.587
y = 3.26

let Ø be the acute angle formed with the x-axis
tanØ = y/x = 3.26/2.587 = ...
Ø = 51.56°

L^2 = 2.587^2 + 3.26^2
..
..
L = appr 4.16

Answer 2

if the line goes through (x,0),

tanθ = 2/(x-1)
so, secθ = √(x^2-2x+5) / (x-1)

the distance AB is

d = x secθ = x√(x^2-2x+5) / (x-1)
dd/dx = (x^3-3x^2+3x-5)/[(x-1)^2 √(x^2-2x+5)]
the denominator is never zero for x>1, so dd/dx=0 when

x^3-3x^2+3x-5 = 0
x = 1+∛4
so,
tanθ = 2/∛4 and θ=51.56°
d = 1+∛4 secθ = 4.16

Or, we could use d as a function of θ

2/(x-1) = tanθ, so x = 1+2cotθ
d = x secθ = (1+2cotθ)secθ = secθ + 2cscθ

dd/dθ = secθtanθ - 2cscθcotθ
= (sin^3θ - 2cos^3θ)/(sin^2θcos^2θ)
dd/dθ = 0 when sinθ = ∛2 cosθ
θ = 51.53°
pretty close to the other answer.

Answer 3

dang! we got the same answer!