The locus of the pole of the chords of the parabola y^2-4ax=0 which subtend right angle at it's vertex is ??

ler the pole be (x_1,y₁)  then polar is yy_1-2a(x+x₁) on homogenising with parabola we get  homogenous pair of straight lines ie           8a²-4ay₁xy+2ax₁y²=0 since it subtend right angle at vertex then coefficient of x²+ coefficient of y² =0 therefore 8a²+2ax₁ =0 therefore locus is x₁+4a=0 , x+4a=0

let the pole be (x1,y1) then polar is yy1-2a(x+x1) on homogenising with parabola we get homogenous pair of straight lines ie 8a2-4ay1xy+2ax1y2=0 since it subtend right angle at vertex then coefficient of x2+ coefficient of y2 =0 therefore 8a2+2ax1 =0 therefore locus is x1+4a=0 , x+4a=0

ler the pole be (x_1,y₁)  then polar is yy_1-2a(x+x₁) on homogenising with parabola we get  homogenous pair of straight lines ie           8a²-4ay₁xy+2ax₁y²=0 since it subtend right angle at vertex then coefficient of x²+ coefficient of y² =0 therefore 8a²+2ax₁ =0 therefore locus is x₁+4a=0 , x+4a=0