Question

If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is:

• A. \( \frac{4}{\sqrt{17}} \)
• B. \( \frac{\sqrt{5}}{16} \)
• C. \( \frac{3}{\sqrt{19}} \)
• D. \( \frac{\sqrt{5}}{7} \)

Hint:

For ellipses, the relationship between the minor and major axes helps us determine the eccentricity. Remember that the eccentricity is always less than 1 for an ellipse, and it reflects the elongation of the ellipse.

Answer 1

The Correct Option is A

The length of the minor axis is one fourth of the distance between the foci. Let the length of the minor axis be \( b \), and the semi-major axis length be \( a \). The distance between the foci is \( 2c \), where \( c \) is the distance from the center to a focus. Given: \[b = \frac{1}{4} (2c) = \frac{c}{2}\] The relationship between \( a \), \( b \), and \( c \) in an ellipse is: \[c^2 = a^2 - b^2\] Substitute \( b = \frac{c}{2} \): \[c^2 = a^2 - \left(\frac{c}{2}\right)^2\] \[c^2 = a^2 - \frac{c^2}{4}\] Multiply by 4: \[4c^2 = 4a^2 - c^2\] \[5c^2 = 4a^2\] \[c^2 = \frac{4a^2}{5}\] The eccentricity \( e \) of an ellipse is: \[e = \sqrt{1 - \frac{b^2}{a^2}}\] Substitute \( b = \frac{c}{2} \): \[e = \sqrt{1 - \frac{\left(\frac{c}{2}\right)^2}{a^2}} = \sqrt{1 - \frac{c^2}{4a^2}}\] Substitute \( c^2 = \frac{4a^2}{5} \): \[e = \sqrt{1 - \frac{\frac{4a^2}{5}}{4a^2}} = \sqrt{1 - \frac{1}{5}}\] \[e = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}\] The eccentricity is \( e = \frac{2}{\sqrt{5}} \).