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

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

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