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

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