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The major axis of a standard ellipse equals the transverse axis of a standard hyperbola. Their director circles have radii 2R and R respectively. If e1 and e2 are the eccentricities of the ellipse and hyperbola, find the correct relation.

1. 4e2² - e1² = 6
2. 4e1² - e2² = 6
3. e1² - 4e2² = 2
4. 2e1² - e2² = 4

Correct answer: 4e2² - e1² = 6

Solution

With common a: ellipse director circle² = a²(2 - e1²) = 4R² and hyperbola director circle² = a²(2 - e2²) = R²; dividing gives (2 - e1²)/(2 - e2²) = 4, leading to 4e2² - e1² = 6.

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