Question

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

पर्याय

• xy = 1

•  2xy + 4x − 4y − 1= 0

• x² − y² = 1

• 2xy − 4x + 4y + 1 = 0

Answer 1

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:

\[(x - 1 )^2 + (y + 1 )^2 = \left( x - y + 1 \right)^2 \]

\[ x^2 - 2x + 1 + y^2 + 1 + 2y = x^2 + y^2 + 1 - 2xy - 2y + 2x\]

\[2xy - 4x + 4y + 1 = 0\]