Question:
<p>The eccentricity of the ellipse <span class="math-tex">\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\)</span> = 1 whose latus rectum is half of its major axis is:</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{\sqrt{2}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\sqrt{\frac{2}{3}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\sqrt{3}}{4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\sqrt{3}}{2}\)</span></p>
Step-by-Step Solution
Key Concept: Equate the latus rectum formula $2b^2/a$ to half the major axis length $2a$ and use the relation $b^2=a^2(1-e^2)$ to solve for eccentricity.
<p>Latus rectum = <span class="math-tex">$2 \frac{b^{2}}{a}=\frac{1}{2}$</span> (2a) (Given)<br />
<span class="math-tex">$\Rightarrow$</span> 2b<sup>2</sup> = a<sup>2</sup><br />
<span class="math-tex">$\Rightarrow$</span> 2a<sup>2</sup>(1 - e<sup>2</sup>) = a<sup>2</sup><br />
<span class="math-tex">$\Rightarrow$</span> 2(1 - e<sup>2</sup>) = 1<br />
<span class="math-tex">$\Rightarrow$</span> 1 - e<sup>2</sup> = <span class="math-tex">$\frac{1}{2}$</span><br />
<span class="math-tex">$\Rightarrow$</span> e<sup>2</sup> = <span class="math-tex">$\frac{1}{2}$</span><br />
<span class="math-tex">$\Rightarrow$</span> e = <span class="math-tex">$\frac{1}{\sqrt{2}}$</span></p>
Correct Answer: A