匿名使用者
匿名使用者 發問時間: 科學數學 · 7 個月前

1. Determine the values of a and b such that the ellipse a (x²/a²)+(y²/b²)=1 contains the circle (x- 1)² + y²= 1 and has least area .?

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  • 匿名使用者
    7 個月前
    最佳解答

     

    Question:

    Determine the values of a and b such that

    the ellipse (x²/a²) + (y²/b²) = 1 contains the circle (x - 1)² + y² = 1 and has least area.

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    Solution:

    Since the ellipse has the least area, the circle touches the ellipse.

    { x²/a² + y²/b² = 1 ...... ①

    { (x - 1)² + y² = 1 ...... ②

    a²b²① - a²②:

    (b²x² + a²y²) - [a²(x - 1)² + a²y²] = a²b² - a²

    (b² - a²)x² + 2a²x - a²b² = 0

    ∵ △ = 0

    4a⁴ + 4a²b²(b² - a²) = 0

    a² + b⁴ - a²b² = 0

    a² = b⁴/(b² - 1)

    Consider the area of the ellipse A = π ab where a, b > 0,

    A = π ab = π b³/√(b² - 1)

    dA/db = π [3b²√(b² - 1) - b⁴/√(b² - 1)]/(b² - 1) = π b²(2b² - 3)/[(b² - 1)√(b² - 1)]

    When dA/db = 0, b = √(3/2) or 0 (rejected) or -√(3/2) (rejected)

    dA/db < 0 when 1 < b < √(3/2)

    dA/db > 0 when b > √(3/2)

    ∴ A obtains the least area when b = √(3/2).

    When b² = 1.5, a² = 1.5²/(1.5 - 1) = 9/2.

    ∵ (-a)² = a² and (-b)² = b²

    ∴ a = ±3√(2)/2, b = ±√(6)/2

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