If the product of eccentricities of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=-1$ is $1$,then $b^2=$

  • A
    $\frac{12}{25}$
  • B
    $144$
  • C
    $25$
  • D
    $\frac{144}{25}$

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