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:
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:
- 5 (Incorrect)
- 7 (Correct)
- 9 (Incorrect)
- 4 (Incorrect)