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Home  >>  EAMCET  >>  Mathematics
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If the foci of the ellipse $\large\frac{x^2}{25}+\frac{y^2}{16}=1$ and the hyperbola $\large\frac{x^2}{4}-\frac{y^2}{b^2}=1$ coincide, then $b^2=$

\[(a)\;4 \quad (b)\;5 \quad (c)\;8 \quad (d)\;9 \]
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(b) 5
answered Nov 7, 2013 by pady_1
 

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