Question

A variable line l passing the point B(2, 5) intersects the lines 2x² − 5xy - 2y² = 0 at P and Q. Find the locus of the point R such that the distances BP, BR and BQ are in harmonic progression (HP).

Answer 1

The given equation 2x² − 5xy + 2y² = 0 represents the pair of lines 2x − y = 0 and x − 2y = 0. Let the equation of the line through B(2, 5) be

x - 2/cosθ = y - 2/sinθ = γ

See Fig. Let BP =  γ₁, BQ =  γ₂ and BR = γ . Therefore

P = (2 + γ₁ cosθ, 5 + γ₁ sinθ)

and Q = (2 + γ₂ cosθ, 5 + γ₂ sinθ)

Since point P lies on x − 2y = 0, we have

 (2 + γ₁ cosθ - 2(5 + γ₁ sinθ) = 0

γ₁  = 8/cosθ - 2 sinθ)  ....(1)

Similarly, Q lies on 2x − y = 0, we have

γ₂  = 1/cosθ- sinθ)  ....(2)

By hypothesis, 

1/γ₁ + 1/γ₂ = 2/γ

Therefore, from Eqs. (1) and (2), we get

The locus of R is 17x − 10y = 0.