To find the value of \( b^2 \) such that the ellipse and hyperbola have coinciding foci, let's analyze each conic section step by step.
1. Ellipse Analysis
The given ellipse equation is: \[ \frac{x^2}{16} + \frac{y^2}{b^2} = 1 \] For an ellipse of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \( a > b \):
Semi-major axis length (\( a \)): \( a = \sqrt{16} = 4 \)
Semi-minor axis length (\( b \)): \( b \) (given)
Distance of each focus from the center (\( c \)): \( c = \sqrt{a^2 - b^2} = \sqrt{16 - b^2} \)
2. Hyperbola Analysis
The given hyperbola equation is: \[ \frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25} \] First, rewrite it in the standard form: \[ \frac{x^2}{\frac{144}{25}} - \frac{y^2}{\frac{81}{25}} = 1 \] Thus: \( a^2 \): \( \frac{144}{25} \) ⇒ \( a = \frac{12}{5} \)
\( b^2 \): \( \frac{81}{25} \) ⇒ \( b = \frac{9}{5} \)
Distance of each focus from the center (\( c \)): \( c = \sqrt{a^2 + b^2} = \sqrt{\frac{144}{25} + \frac{81}{25}} = \sqrt{\frac{225}{25}} = \sqrt{9} = 3 \)
3. Equating the Foci
Since the foci of the ellipse and hyperbola coincide: \[ \sqrt{16 - b^2} = 3 \] Square both sides to solve for \( b^2 \): \[ 16 - b^2 = 9 \\ b^2 = 16 - 9 \\ b^2 = 7 \] Final Answer \( 7 \)