engineering mathematics MCQ #571

For hyperbola x²/9 - y²/16 = 1, foci are at:

engineering mathematics MCQ #571

  1. Question 1

    Q1. For hyperbola x²/9 - y²/16 = 1, foci are at:

    • A) (±5, 0)
    • B) (±4, 0)
    • C) (±3, 0)
    • D) (±2, 0)

    Answer: (±5, 0)

    Explanation: c² = a² + b² → c = √(9 + 16) = 5 → foci (±5, 0).