Q: The equation of a line passing through the centre of a rectangular hyperbola is x – y – 1 = 0. If one of its asymptotes is 3x – 4y – 6 = 0, the equation of the other asymptote is

(A) 4x – 3y + 17 = 0

(B) -4x – 3y + 17 = 0

(C) -4x + 3y + 1 = 0

(D) 4x + 3y + 17 = 0

Sol. We know that asymptotes of rectangular hyperbola are mutually perpendicular, thus other asymptote should be 4x + 3y + λ = 0

Intersection point of asymptotes is also the centre of the hyperbola.

Hence intersection point of 4x +3y + λ = 0 and 3x -4y -6 = 0 should lie on the line x- y -1= 0, using it λ can be easily obtained.

Hence (D) is the correct answer.