The correct option is A xcotα−y=0, xtanα+y=0
Given, x2+2xycot2α−y2=a2
⇒x2+2xy (1−tan2α)2tanα−y2=a2
⇒x2+xy(cotα−tanα)−y2=a2
For asymptotes
x2+xy(cotα−tanα)−y2+λ=0
this will be a pair for st. line if
Δ=0⇒λ=0
⇒x(x+ycotα)−ytanα(x+ycotα)=0⇒(x−ytanα)(x+ycotα)=0
∴ separate equation asymptotes are
x+ycotα=0;x−ytanα=0