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A hyperbola has the same foci as the ellipse x²/25 + y²/9 = 1. If the hyperbola has eccentricity 2, find its equation.
- x²/4 - y²/12 = 1
- x²/12 - y²/4 = 1
- 3x² - y² + 12 = 0
- 9x² - 25y² - 225 = 0
Correct answer: x²/4 - y²/12 = 1
Solution
Ellipse foci at (+/-4, 0); hyperbola has c = 4 and e = 2 so a = c/e = 2, b² = c² - a² = 12, giving x²/4 - y²/12 = 1.
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