If the length of the minor axis of an ellipse is equal to one fourth of the distance betwe…

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 :

Accepted Answer

The length of the minor axis is equal to one-fourth of the distance between the foci. Mathematically, this can be expressed as:

$ 2b = \frac{1}{4}(2ae) $

This simplifies to:

$ b = \frac{ae}{4} $

Given that the relationship between $ b $, $ a $, and $ e $ is:

$ \frac{b^2}{a^2} = 1 - e^2 $

Substitute $ b = \frac{ae}{4} $ into the equation:

$ \left(\frac{ae}{4}\right)^2 = a^2 \cdot \left(1 - e^2\right) $

Expanding and simplifying gives:

$ \frac{a^2 e^2}{16} = a^2(1 - e^2) $

Divide both sides by $ a^2 $:

$ \frac{e^2}{16} = 1 - e^2 $

Rearrange and solve for $ e^2 $:

$ \frac{e^2}{16} + e^2 = 1 $

$ \frac{17e^2}{16} = 1 $

Solve for $ e $:

$ e^2 = \frac{16}{17} $

$ e = \frac{4}{\sqrt{17}} $

Thus, the eccentricity of the ellipse is:

$\frac{4}{\sqrt{17}}$

Solutions · 1