Question

The vertices of the hyperbola 9x² – 16y² – 36x + 96y – 252 = 0 are 

(a) (6, 3), (– 6, 3)

(b) (– 6, 3), (– 6, – 3)

(c) (6, – 3), (2, – 3)

(d) (6, 3), (– 2, 3)

Answer 1

The correct option (d) (6, 3), (– 2, 3)

Explanation:

hyperbola: 9x² – 16y² – 36x + 96y – 252 = 0

∴    9(x² – 4x + 4) – 16(y² – 6y + 9) = 252 + 36 – 144

∴     9(x – 2)² – 16(y – 3)² = 144

∴    [(x – 2)²/16] – [(y – 3)²/9] = 1

∴  (X²/A²) – (Y²/B²) = 1

∴    X = x – 2  and     Y = y – 3

∴ centre is at (2, 3). as a = 4 ⇒ distance from centre to vertex is 4 units.

⇒ x = 6, – 2 and y – 3 = 0 ⇒ y = 3

∴ vertices (6, 3) and (– 2, 3).