Question:
<p>If the foci of the ellipse <span class="math-tex">\(\frac{x^2}{25}+\frac{y^2}{b^2}=1\)</span> and the hyperbola <span class="math-tex">\(\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}\)</span> coincide, then the value of <span class="math-tex">\(b^2\)</span> is.</p>
<p style="display:inline">9</p>
<p style="display:inline">12</p>
<p style="display:inline">16</p>
<p style="display:inline">3</p>
Step-by-Step Solution
Key Concept: When the foci of an ellipse and hyperbola coincide, the squared focal distance $a^2e^2$ is equal for both, calculated as $a^2-b^2$ for the ellipse and $a^2+b^2$ for the hyperbola.
<p>If eccentricities of ellipse and hyperbola are <span class="math-tex">$e$</span> and <span class="math-tex">$e_1$</span><br />
<span class="math-tex">$ \therefore \text { foci }( \pm a e, 0) \text { and }\left( \pm a_1 e_1, 0\right)$</span><br />
Hence <span class="math-tex">$a^2 e^2=a_1{ }^2 e_1^2$</span><br />
<span class="math-tex">$ \Rightarrow a^2\left(1-\frac{b^2}{a^2}\right)=a_1^2\left(1+\frac{b_1^2}{a_1^2}\right) $</span><br />
<span class="math-tex">$ \Rightarrow a^2-b^2=a_1^2+b_1^2 $</span><br />
<span class="math-tex">$ \Rightarrow 25-b^2=\frac{144}{25}+\frac{81}{25}=9 $</span><br />
<span class="math-tex">$ \Rightarrow b^2=16 $</span></p>
Correct Answer: C