engineering mathematics MCQ #559

For hyperbola x²/16 - y²/9 = 1, the directrices are at x = ±:

engineering mathematics MCQ #559

  1. Question 1

    Q1. For hyperbola x²/16 - y²/9 = 1, the directrices are at x = ±:

    • A) 16/5
    • B) 5/16
    • C) 4/5
    • D) 5/4

    Answer: 16/5

    Explanation: Directrices of hyperbola at x = ±a/e. e = c/a = √(a² + b²)/a = 5/4 ⇒ x = 16/5.