Aspire Faculty ID #10198 · Topic: NIMCET 2015 · Just now
NIMCET 2015

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

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