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

Question

The equation of the conic with focus at (1, -1) directrix along x - y + 1 = 0 and with eccentricity √2 is (A) x2 - y2 = 1 (B) xy = 1 (C) 2xy - 4x + 4y + 1 = 0 (D) 2xy + 4x - 4y - 1 = 0

Tardigrade Admin · Accepted Answer

Let P (x, y) be any point on the conic. Then, √(x -1)2 + ( y + 1)2 = √2 ((x - y + 1 /√2)) ⇒ (x - 1)2 + ( y +1)2 = (x - y + 1)2 ⇒ 2xy - 4 x + 4y + 1 = 0

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

$x^{2} - y^{2} = 1$

$x y = 1$

$2 x y - 4 x + 4 y + 1 = 0$

$2 x y + 4 x - 4 y - 1 = 0$

Let $P (x, y)$ be any point on the conic. Then,
$(x - 1)^{2} + (y + 1)^{2} = 2 (\frac{x - y + 1}{2})$
$\Rightarrow (x - 1)^{2} + (y + 1)^{2} = (x - y + 1)^{2}$
$\Rightarrow 2 x y - 4 x + 4 y + 1 = 0$

Q. The equation of the conic with focus at (1,−1) directrix along x−y+1=0 and with eccentricity 2​ is

x2−y2=1

xy=1

2xy−4x+4y+1=0

2xy+4x−4y−1=0