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For the ellipse x²/a² + y²/b² = 1 (a > b), one extremity of the minor axis together with the two foci forms an equilateral triangle. What is the eccentricity of the ellipse?
- 1/2
- 1/sqrt(2)
- sqrt(5)/3
- sqrt(7)/4
Correct answer: 1/2
Solution
The distance from a focus (a*e, 0) to the minor-axis tip (0, b) simplifies to a; equating it to the focal separation 2*a*e gives e = 1/2.
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