Locus of the poles of focal chord of a parabola is

Step-by-step solution

Equation of P Q is aligned y-2 a t₁ & = 2 a t₂-2 a t₁ a t₂^2-a t₁^2 (x-a t₁^2 ) (t₂+t₁ ) (y-2 a t₁ ) & =2 (x-a t₁^2 ) (t₁+t₂ ) y-2 x & =2 a t₁ t₂ aligned This line is passing through (a, 0) aligned & & (t₁+t₂ ) (-2 a t₁ )= 2 t₁+t₂ (a-a t₁^2 ) & & t₁ t₂=-1 aligned Let P (x₁, y₁ ) be the pole of (i) w.r.t. y^2=4 a x Its polar is y y₁=2 a (x+x₁ ) From equaiton (i) and (iii), we get t₁+t₂ y₁ = 1 a = 2 a t₁ t₂ 2 a x₁ From last two relations, we get array ll & x₁=a t₁ t₂ & x₁=-a locus is x=-a & array