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Let a and c be positive real numbers. If the ellipse x²/(4c²) + y²/c² = 1 has four distinct points in common with the circle x² + y² = 9a², then which inequality must hold?

1. 6ac + 9a² - 2c² > 0
2. 6ac + 9a² - 2c² < 0
3. 9ac - 9a² - 2c² < 0
4. 9ac - 9a² - 2c² > 0

Correct answer: 9ac - 9a² - 2c² > 0

Solution

Four common points require the circle radius 3a to lie between the semi-minor (c) and semi-major (2c) axes: c < 3a < 2c, equivalently 1.5a < c < 3a. This band is exactly 2c² - 9ac + 9a² < 0, i.e. 9ac - 9a² - 2c² > 0.

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