The correct option is C y+1=0
Let P be any point on hyperbola and F be the focus and S be foot of perpendicular on directrix from hyperbola.
As we know that,
PF2=e2PS2
x2+y2=2(x+y+1√2)2
Solving it we get hyperbola equation as, 2x+2y+2xy+1=0
Now combined equation of pair of asymptotes is
2x+2y+2xy+1+λ=0
As it represents pair of straight lines,
abc+2fgh−af2−bg2−ch2=0
a=0, b=0, h=1, g=1, f=1, c=1+λ
⇒ 2×1×1×1−(1+λ)=0
∴ λ=1
So , combined equation of asymptotes are
2x+2y+2xy+2=0
⇒ (2x+2)(y+1)=0
∴ x+1=0 and y+1=0 are the asymptote equations