BITSAT 2012 Paper

We have x2–y2 – 4x + 4y + 16 = 0
⇒ (x2 – 4x) – (y2 – 4y) = 16
⇒ (x2 – 4x + 4) – (y2 – 4y + 4) = – 16
⇒ (x–2)2–(y–2)2 = – 16
⇒

(x−2)2

42

−

(y−2)2

42

= 1
This is rectangular hyperbola, whose eccentricity is always √2