engineering mathematics MCQ #590

Equation of hyperbola with vertices (±3, 0) and foci (±5, 0) is:

engineering mathematics MCQ #590

  1. Question 1

    Q1. Equation of hyperbola with vertices (±3, 0) and foci (±5, 0) is:

    • A) x²/9 - y²/16 = 1
    • B) x²/16 - y²/9 = 1
    • C) x²/25 - y²/16 = 1
    • D) x²/16 - y²/25 = 1

    Answer: x²/9 - y²/16 = 1

    Explanation: a = 3, c = 5 ⇒ b² = c² - a² = 25 - 9 = 16.