engineering mathematics MCQ #472

The foci of the hyperbola 9x² - 16y² = 144 are

engineering mathematics MCQ #472

  1. Question 1

    Q1. The foci of the hyperbola 9x² - 16y² = 144 are

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

    Answer: (±5, 0)

    Explanation: For 9x² - 16y² = 144, we have x²/16 - y²/9 = 1, so a² = 16, b² = 9, and c² = a² + b² = 25, hence c = 5.