Question

The equation of the conic with focus at (1, −1) directrix along x − y + 1 = 0 and eccentricity $\sqrt{2}$ is
(a) xy = 1
(b) 2xy + 4x − 4y − 1= 0
(c) x² − y² = 1
(d) 2xy − 4x + 4y + 1 = 0

Answer 1

(d) 2xy − 4x + 4y + 1 = 0

Let $P(x,y)$ be any point on the hyperbola.
Then, the distance of any point from the focus is eccentricity times the distance from the directrix.

$\therefore \sqrt{{\left(x-1\right)}^2+{\left(y+1\right)}^2}=\sqrt{2}\left|\frac{x-y+1}{\sqrt{2}}\right|$

Squaring both the sides, we get:
$$\begin{gathered} (x-1{)}^2+(y+1{)}^2={\left(x-y+1\right)}^2 \\ {x}^2-2x+1+y^2+1+2y=x^2+y^2+1-2xy-2y+2x \\ 2xy-4x+4y+1=0 \end{gathered}$$