The value of $b^2$ such that the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$ and the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ coincide is

  • A
    $1$
  • B
    $5$
  • C
    $7$
  • D
    $9$

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