Question:
<p>If the latus rectum of an ellipse be equal to half of its minor axis, then its eccentricity is:</p>
<p style="display:inline"><span class="math-tex">\(\frac {\sqrt 2}3\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac {\sqrt 3}2\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac 32\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac 23\)</span></p>
Step-by-Step Solution
Key Concept: Determine the eccentricity by using the given geometric relationship to find the ratio of the semi-minor axis to the semi-major axis.
<p><span class="math-tex">$\frac{2 b^{2}}{a}$</span> = b<br />
<span class="math-tex">$\Rightarrow \frac{b}{a}=\frac{1}{2}$</span><br />
<span class="math-tex">$\Rightarrow\frac{b^{2}}{a^{2}}=\frac{1}{4}$</span><br />
Hence e = <span class="math-tex">$\sqrt{1-\frac{b^{2}}{a^{2}}}=\frac{\sqrt{3}}{2}$</span></p>
Correct Answer: B