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Step 1: Identify the Standard Forms
The ellipse is given by $ \frac{x^2}{16} + \frac{y^2}{b^2} = 1 $. This corresponds to the standard form
$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ with $a^2 = 16$, so $a = 4$.
The hyperbola is given by $ \frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25} $. We can rewrite it as
$ \frac{x^2}{(144/25)} - \frac{y^2}{(81/25)} = 1 $.
Thus for the hyperbola’s standard form
$ \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 $,
we have $A^2 = \frac{144}{25}$ and $B^2 = \frac{81}{25}$.
Step 2: Find the Foci of the Hyperbola
For a hyperbola of the form $ \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 $, the eccentricity $ e $ is given by
$ e = \sqrt{1 + \frac{B^2}{A^2}}. $
Here,
$ A^2 = \frac{144}{25}, \quad B^2 = \frac{81}{25}. $
Thus,
$ e = \sqrt{1 + \frac{81/25}{144/25}} = \sqrt{1 + \frac{81}{144}} = \sqrt{\frac{144 + 81}{144}} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}. $
The foci of this hyperbola lie at $ (\pm Ae,\,0) $, so
$ A = \frac{12}{5}, \quad Ae = \frac{12}{5} \times \frac{5}{4} = 3. $
Hence, the foci of the hyperbola are $ ( \pm 3, 0 ). $
Step 3: Equate the Foci of the Ellipse and Hyperbola
Since the foci coincide, the ellipse must also have foci at $ ( \pm 3, 0 ). $ For an ellipse of the form
$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ with $a = 4$, the eccentricity is
$ e = \frac{c}{a} $ where $ c $ is the focal distance from the origin.
Because the ellipse’s foci are at $ ( \pm 3,0 ) $, we have $ c = 3 $. Therefore:
$ \frac{c}{a} = e = \frac{3}{4}. $
Step 4: Find $ b^2 $ for the Ellipse
For an ellipse, $ b^2 = a^2(1 - e^2). $
Given $ a = 4 $ and $ e = \frac{3}{4} $, we find
$ e^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16}, $
Thus,
$ b^2 = 16 \bigl(1 - \frac{9}{16}\bigr) = 16 \cdot \frac{7}{16} = 7. $
Answer
$ b^2 = 7. $
