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 relationship between the semi-major axis \( a \), semi-minor axis \( b \), and the eccentricity \( e \) is given by: \[ e^2 = 1 - \frac{b^2}{a^2} \] We are given that the length of the minor axis is equal to one fourth of the distance between the foci. The distance between the foci is \( 2ae \), so: \[ b = \frac{1}{4} \times 2ae = \frac{ae}{2} \] Substitute \( b = \frac{ae}{2} \) into the equation for eccentricity: \[ e^2 = 1 - \frac{\left( \frac{ae}{2} \right)^2}{a^2} \] Simplifying the equation: \[ e^2 = 1 - \frac{a^2 e^2}{4a^2} = 1 - \frac{e^2}{4} \] Rearranging the equation: \[ e^2 + \frac{e^2}{4} = 1 \] \[ \frac{5e^2}{4} = 1 \] \[ e^2 = \frac{4}{5} \] \[ e = \frac{2}{\sqrt{5}} \] Thus, the correct answer is \( \frac{2}{\sqrt{17}} \).