engineering mathematics MCQ #402

The foci of the hyperbola x²/16 - y²/9 = 1 are

engineering mathematics MCQ #402

  1. Question 1

    Q1. The foci of the hyperbola x²/16 - y²/9 = 1 are

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

    Answer: (±5, 0)

    Explanation: For hyperbola x²/a² - y²/b² = 1, foci are (±√(a² + b²), 0). Here, a² = 16, b² = 9, so √(a² + b²) = √25 = 5.