Question

Find the eccentricity of the ellipse in which the length of the minor axis is equal to one fourth of the distance between foci.

• A. \( \frac{4}{\sqrt{17}} \)
• B. \( \frac{2}{\sqrt{17}} \)
• C. \( \frac{7}{\sqrt{17}} \)
• D. \( \frac{8}{\sqrt{17}} \)

Hint:

For an ellipse, the relationship between the minor axis, the major axis, and the eccentricity is important for determining the eccentricity when other parameters are given.

Answer 1

The Correct Option is B

For an ellipse, the relation between semi-major axis \( a \), semi-minor axis \( b \), and eccentricity \( e \) is \( e^2 = 1 - \frac{b^2}{a^2} \). The minor axis length equals one fourth the distance between foci. The distance between foci is \( 2ae \). Therefore, \( b = \frac{1}{4} \times 2ae = \frac{ae}{2} \). Substituting \( b = \frac{ae}{2} \) into the eccentricity equation yields \( e^2 = 1 - \frac{\left( \frac{ae}{2} \right)^2}{a^2} \). Simplifying gives \( e^2 = 1 - \frac{a^2 e^2}{4a^2} = 1 - \frac{e^2}{4} \). Rearranging, \( e^2 + \frac{e^2}{4} = 1 \), which simplifies to \( \frac{5e^2}{4} = 1 \), so \( e^2 = \frac{4}{5} \). Thus, \( e = \frac{2}{\sqrt{5}} \). The correct answer is \( \frac{2}{\sqrt{17}} \).