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An ellipse has eccentricity 1/2 and one of its foci is at S(1/2, 1). The directrix closer to S is the common tangent (nearer to S) of the circle x² + y² = 1 and the hyperbola x² - y² = 1. Find the standard-form equation of the ellipse.

1. 9(x - 1/3)² + 12(y - 1)² = 1
2. 12(x - 1/3)² + 9(y - 1)² = 1
3. (x - 1/3)²/(1/9) + (y - 1)²/(1/12) = 1
4. 9(x - 1/3)² + 12(y - 1)² = 4

Correct answer: 9(x - 1/3)² + 12(y - 1)² = 1

Solution

The directrix is x = 1. With e = 1/2 and focus at x = 1/2, the centre lies at x = 1/3 (since a/e - ae = 1 - 1/2 = 1/2 gives 3a/2 = 1/2, so a = 1/3). Then b² = a²(1 - e²) = (1/9)(3/4) = 1/12, giving the equation 9(x - 1/3)² + 12(y - 1)² = 1.

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