Final answer:
The eccentricity of an ellipse with the distance between the directrices being three times the distance between the foci is 1/√3. The eccentricity is derived from the relationship between the distance of the foci and the semimajor axis of the ellipse, giving us the calculation e = f/a.
Step-by-step explanation:
The student is asking a question about the eccentricity of an ellipse. Given that the distance between the directrices is three times the distance between the foci, we can find the eccentricity through an understanding of the relationship between these distances and the ellipse's semimajor and semiminor axes. For an ellipse, eccentricity e is defined as the ratio of the distance f from the center to a focus, to the semimajor axis a: e = f/a. The sum of the distances from the foci to any point on the ellipse is 2a, which is a constant and equal to the length of the major axis.
Since the distance between the directrices is three times the distance between the foci, and the directrices are at a distance a/e from the center (where a is the semimajor axis and e the eccentricity), we have 2a/e = 3(2f). This simplifies to a/e = 3f, which means e = a/3f. But since e = f/a, we have f/a = a/3f, resulting in f2 = a2/3. Taking square roots, f = a/√3, and hence the eccentricity e = f/a = 1/√3. Thus the correct answer is b, which is 1/√3.