If the foci of the ellipse \( \frac{x^2}{16}+\frac{y^2}{b^2}=1 \) and the foci of the hyperbola \( \frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25} \) coincide, then the value of \( b^2 \) is
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For ellipse, use \( c^2=a^2-b^2 \), while for hyperbola, use \( c^2=a^2+b^2 \)Always convert equation to standard form first.