Question:
<p>Let <span class="math-tex">\(e_{1}\)</span> be the eccentricity of the hyperbola <span class="math-tex">\(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)</span> and <span class="math-tex">\(e_{2}\)</span> be the eccentricity of the ellipse <span class="math-tex">\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)</span>, <span class="math-tex">\(a \gt b\)</span>, which passes through the foci of the hyperbola. If <span class="math-tex">\(e_{1} e_{2}=1\)</span>, then the length of the chord of the ellipse parallel to the <span class="math-tex">\(x\)</span>-axis and passing through <span class="math-tex">\((0,2)\)</span> is:</p>
<p style="display:inline"><span class="math-tex">\(3 \sqrt{5}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{10 \sqrt{5}}{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(4 \sqrt{5}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{8 \sqrt{5}}{3}\)</span></p>
Step-by-Step Solution
<p>Hyperbola: <span class="math-tex">$\frac{x^{2}}{5}-\frac{y^{2}}{9}=1$</span><br />
<span class="math-tex">$e_{1}=\frac{5}{4}$</span><br />
<span class="math-tex">$\therefore e_{1} e_{2}=1 \Rightarrow e_{2}=\frac{4}{5}$</span><br />
Also, ellipse is passing through <span class="math-tex">$( \pm 5,0)$</span><br />
<span class="math-tex">$\therefore a=5$</span> and <span class="math-tex">$b=3$</span><br />
Ellipse: <span class="math-tex">$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$</span><br />
<img src="https://media-mycbseguide.s3.amazonaws.com/images/question_images/1775195486-6xf52j.jpg" style="height:140px; width:200px" /><br />
End point of chord are <span class="math-tex">$\left( \pm \frac{5 \sqrt{5}}{3}, 2\right)$</span><br />
<span class="math-tex">$\therefore L_{P Q}=\frac{10 \sqrt{5}}{3}$</span></p>
Correct Answer: B