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 :
Solution
<p>The length of the minor axis is equal to one-fourth of the distance between the foci. Mathematically, this can be expressed as:</p>
<p>$ 2b = \frac{1}{4}(2ae) $</p>
<p>This simplifies to:</p>
<p>$ b = \frac{ae}{4} $</p>
<p>Given that the relationship between $ b $, $ a $, and $ e $ is:</p>
<p>$ \frac{b^2}{a^2} = 1 - e^2 $</p>
<p>Substitute $ b = \frac{ae}{4} $ into the equation:</p>
<p>$ \left(\frac{ae}{4}\right)^2 = a^2 \cdot \left(1 - e^2\right) $</p>
<p>Expanding and simplifying gives:</p>
<p>$ \frac{a^2 e^2}{16} = a^2(1 - e^2) $</p>
<p>Divide both sides by $ a^2 $:</p>
<p>$ \frac{e^2}{16} = 1 - e^2 $</p>
<p>Rearrange and solve for $ e^2 $:</p>
<p>$ \frac{e^2}{16} + e^2 = 1 $</p>
<p>$ \frac{17e^2}{16} = 1 $</p>
<p>Solve for $ e $:</p>
<p>$ e^2 = \frac{16}{17} $</p>
<p>$ e = \frac{4}{\sqrt{17}} $</p>
<p>Thus, the eccentricity of the ellipse is:</p>
<p>$\frac{4}{\sqrt{17}}$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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