Step-by-step Solution:

1. Given Equations: - Ellipse: \[ b^2x^2 + 16y^2 = 16b^2 \] - Hyperbola: \[ 81x^2 - 144y^2 = \frac{81 \times 144}{25} \] 2. Standard Form of Ellipse: Divide the ellipse equation by \( 16b^2 \): \[ \frac{x^2}{16} + \frac{y^2}{b^2} = 1 \] For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the distance of each focus from the center is: \[ c = \sqrt{a^2 - b^2} \] Here, \( a^2 = 16 \) and \( b^2 = b^2 \), so: \[ c = \sqrt{16 - b^2} \] 3. Standard Form of Hyperbola: Divide the hyperbola equation by \( \frac{81 \times 144}{25} \): \[ \frac{x^2}{\frac{144}{25}} - \frac{y^2}{\frac{81}{25}} = 1 \] For a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the distance of each focus from the center is: \[ c = \sqrt{a^2 + b^2} \] Here, \( a^2 = \frac{144}{25} \) and \( b^2 = \frac{81}{25} \), so: \[ c = \sqrt{\frac{144}{25} + \frac{81}{25}} = \sqrt{\frac{225}{25}} = \sqrt{9} = 3 \] 4. Set Foci Equal: Since the foci coincide: \[ \sqrt{16 - b^2} = 3 \] Square both sides: \[ 16 - b^2 = 9 \] Solve for \( b^2 \): \[ b^2 = 16 - 9 = 7 \] \[ b = \sqrt{7} \] Therefore, the value of \( b \) is: \[ \boxed{\sqrt{7}} \]