<p>The foci of the ellipse \(\dfrac{x^2}{16} + \dfrac{y^2}{b^2} = 1\) and the hyperbola \(\dfrac{x^2}{144} - \dfrac{y^2}{81} = \dfrac{1}{25}\) coincide. Then the value of \(b^2\) is</p>
Step-by-Step Solution
Key Concept: First convert the hyperbola to standard form by dividing through by 1/25, then use the fact that foci coincide: both curves share the same c value, where c² = a² - b² for ellipse and c² = a² + b² for hyperbola.
<p><strong>Step 1:</strong> Convert hyperbola to standard form.</p><p>Given: $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$</p><p>Multiply by 25: $\frac{25x^2}{144} - \frac{25y^2}{81} = 1$</p><p>This gives: $\frac{x^2}{144/25} - \frac{y^2}{81/25} = 1$</p><p><strong>Step 2:</strong> Find c for hyperbola.</p><p>For hyperbola: $a_h^2 = \frac{144}{25}$, $b_h^2 = \frac{81}{25}$</p><p>$c_h^2 = a_h^2 + b_h^2 = \frac{144}{25} + \frac{81}{25} = \frac{225}{25} = 9$</p><p>So $c_h = 3$</p><p><strong>Step 3:</strong> Find b² for ellipse using coinciding foci.</p><p>For ellipse: $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ with $a_e^2 = 16$</p><p>Since foci coincide: $c_e = c_h = 3$</p><p>$c_e^2 = a_e^2 - b^2 = 16 - b^2 = 9$</p><p>$b^2 = 16 - 9 = 7$</p><p>∴ Answer: C</p>
Correct Answer: C