Question:
<p>An ellipse has OB as a semi-minor axis, F, F' as its foci and the angle <span class="math-tex">\(\angle\)</span>FBF' is a right angle. Then, the eccentricity of the ellipse is</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{\sqrt{2}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\sqrt{3}}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{\sqrt{3}}\)</span></p>
Step-by-Step Solution
Key Concept: Apply the Pythagorean theorem to the right triangle formed by the foci and the minor axis endpoint, utilizing the property that the distance from a focus to any endpoint of the minor axis equals the semi-major axis length.
<p>We have, <span class="math-tex">\(\angle \mathrm{FBF}^{\prime}=\frac{\pi}{2}\)</span><br />
<span class="math-tex">\(\therefore\)</span> FF'<sup>2</sup> = FB<sup>2</sup> + F'B<sup>2</sup><br />
<span class="math-tex">\(\Rightarrow\)</span> (2ae)<sup>2</sup> = a<sup>2</sup> + a<sup>2</sup> ...<span class="math-tex">\(\because\)</span> x - coordinate of B is zero, <span class="math-tex">\(\therefore\)</span> FB = a - e <span class="math-tex">\(\times\)</span> 0 and F'B = a + e <span class="math-tex">\(\times\)</span> 0]<br />
<span class="math-tex">\(\Rightarrow\)</span> 4e<sup>2</sup> = 2<br />
<span class="math-tex">\(\Rightarrow e=\frac{1}{\sqrt{2}}\)</span></p>
Correct Answer: A