Question

The equation of the conic with focus at $(1,-1)$ directrix along $x-y+1=0$ and with eccentricity $\sqrt{2}$ is

• A. $x^2-y^2=1$
• B. $xy=1$
• C. $2xy-4x+4y+1=0$
• D. $2xy+4x-4y-1=0$

Answer 1

Answer: (C)

From the definition of conic; If $P(x,y)$ is the point on a conic then ratio of its distance from focus to its distance from directrix is a fixed ratio e, called eccentricity. Here focus is $(1,-1)$ and directrix is $x-y+1=0$.
Distance of this point from focus $=\sqrt{(x-1{)}^2+(y+1{)}^2}$
Distance of this point from directrix. $=\left|\frac{x-y+1}{\sqrt{1^2+(-1{)}^2}}\right|$
So, from the definition of conic
$\sqrt{(x-1{)}^2+(y+1{)}^2}=e.\left|\frac{x-y+1}{\sqrt{2}}\right|...(i)$
Squaring both sides of equation (i), we get
$(x-1{)}^2+(y+1{)}^2=e^2\cdot \frac{(x-y+1{)}^2}{2}=(\sqrt{2}{)}^2\frac{(x-y+1{)}^2}{2}$
$=(x-y+1{)}^2$
$\Rightarrow (x-1{)}^2+(y+1{)}^2=(x-y+1{)}^2$
$\Rightarrow x^2-2x+1+y^2+2y+1=x^2-2xy+y^2+2x-2y+1$
$\Rightarrow 2xy-4x+4y+1=0$