Question

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

Answer 1

Answer: (D)

4x + 3y + 17 = 0

Answer 2

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, so that λ = 17