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Question

A hyperbola has focus at origin, its eccentricity is 2 and corresponding directrix is x+y+1=0. The equation of its asymptotes is/are:

A
x+1=0
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B
x1=0
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C
y+1=0
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D
y1=0
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Solution

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+12)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+2fghaf2bg2ch2=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

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