# 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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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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