<p>The foci of the ellipse \(\frac{x^2}{16} + \frac{y^2}{b^2} = 1\) and the hyperbola \(\frac{25x^2}{144} - \frac{25y^2}{81} = 1\) coincide. Then the value of \(b^2\) is:</p>
Step-by-Step Solution
Key Concept: For the ellipse and hyperbola to have coinciding foci, calculate c² for each curve separately (c² = a² - b² for ellipse, c² = a² + b² for hyperbola) and equate them since both must lie on the same axis.
<p><strong>Step 1: Identify ellipse parameters</strong></p><p>Ellipse: $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ with $a^2 = 16$</p><p>Since $a^2 = 16 > b^2$, the major axis is along x-axis.</p><p>For ellipse: $c^2 = a^2 - b^2 = 16 - b^2$</p><p><strong>Step 2: Simplify hyperbola equation</strong></p><p>Hyperbola: $\frac{25x^2}{144} - \frac{25y^2}{81} = 1$</p><p>Divide by 25: $\frac{x^2}{144/25} - \frac{y^2}{81/25} = 1$</p><p>So $A^2 = \frac{144}{25}$ and $B^2 = \frac{81}{25}$</p><p><strong>Step 3: Calculate c² for hyperbola</strong></p><p>For hyperbola with transverse axis along x: $c_h^2 = A^2 + B^2 = \frac{144}{25} + \frac{81}{25} = \frac{225}{25} = 9$</p><p><strong>Step 4: Equate the foci</strong></p><p>Since foci coincide: $c_e^2 = c_h^2$</p><p>$16 - b^2 = 9$</p><p>$b^2 = 7$</p><p>∴ Answer: C</p>
Correct Answer: C