Question:
<p>For some <span class="math-tex">\(\theta \in\left(0, \frac{\pi}{2}\right)\)</span>, if the eccentricity of the hyperbola, x<sup>2</sup> - y<sup>2</sup> sec<sup>2</sup> <span class="math-tex">\(\theta\)</span> = 10 is <span class="math-tex">\(\sqrt{5}\)</span> times the eccentricity of the ellipse, x<sup>2</sup> sec<sup>2</sup><span class="math-tex">\(\theta\)</span> + y<sup>2</sup> = 5, then the length of the latus rectum of the ellipse, is:</p>
<p style="display:inline"><span class="math-tex">\(\sqrt{30}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{2 \sqrt{5}}{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(2 \sqrt{6}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{4 \sqrt{5}}{3}\)</span></p>
Step-by-Step Solution
Key Concept: Standardize the conic equations to relate their eccentricities and identify the ellipse's orientation to correctly calculate its latus rectum.
<p>Hyperbola : <span class="math-tex">\(\frac{x^{2}}{10}-\frac{y^{2}}{10 \cos ^{2} \theta}=1 \Rightarrow e_{1}=\sqrt{1+\cos ^{2} \theta}\)</span> and Ellipse : <span class="math-tex">\(\frac{x^{2}}{5 \cos ^{2} \theta}+\frac{y^{2}}{5}=1\)</span><br />
<span class="math-tex">\(\Rightarrow e_{2}=\sqrt{1-\cos ^{2} \theta}=\sin \theta\)</span><br />
According to the question, <span class="math-tex">\(e_{1}=\sqrt{5} e_{2}\)</span><br />
<span class="math-tex">\(\Rightarrow 1+\cos ^{2} \theta=5 \sin ^{2} \theta \Rightarrow \cos ^{2} \theta=\frac{2}{3}\)</span><br />
Now length of latus rectum of ellipse <span class="math-tex">\(=\frac{2 a^{2}}{b}=\frac{10 \cos ^{2} \theta}{\sqrt{5}}=\frac{20}{3 \sqrt{5}}=\frac{4 \sqrt{5}}{3}\)</span></p>
Correct Answer: D