Question

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

• A. $x+1=0$
• B. $x-1=0$
• C. $y+1=0$
• D. $y-1=0$

Answer 1

Answer: (A), (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,
$P{F}^2=e^{2}P{S}^2$
$x^2+y^2=2{\left(\frac{x+y+1}{\sqrt{2}}\right)}^2$
Solving it we get hyperbola equation as, $2x+2y+2xy+1=0$
Now combined equation of pair of asymptotes is
$2x+2y+2xy+1+\lambda =0$
As it represents pair of straight lines,
$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$
$a=0,b=0,h=1,g=1,f=1,c=1+\lambda$
$\Rightarrow 2\times 1\times 1\times 1-(1+\lambda )=0$
$\therefore \lambda =1$
So , combined equation of asymptotes are
$2x+2y+2xy+2=0$
$\Rightarrow (2x+2)(y+1)=0$
$\therefore x+1=0$ and $y+1=0$ are the asymptote equations