Q.250. Maxima/minima in geometry

Question 250.
Refer to the figure as under.

Lines OA and OB make an angle of 45 degrees.  O (0, 0) is the origin, point P = (43/120, 1/5), and point Q = (37/48, 7/12). A line is drawn through Q, intersecting two lines at points A and B. Lines PA and PB are drawn from P to A and B. Depending on what angle line AB is drawn through Q, total length of AP + PB varies. What is the minimum total length of AP + PB?

Answer 250.
Let A = (x, 0) and B = (b, b)
A, Q, B are collinear
=> (7/12)/(37/48 - x) = b/(b - x)
=> (7/12) (b - x) = b (37/48 - x)
=> b = 28x / (48x - 9)

=> PA + PB
= √[(x - 43/120)^2 + (1/5)^2] + √[(28x/(48x-9) - 43/120)^2 + (28x/(48x-9) - 1/5)^2]

Minimum is ≈ 1.125 at x ≈ 0.625
(as worked out using Wolfram Alpha link as under.)

Wolfram Alpha Link
Link to YA!