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The foci of the ellipse $\large\frac{x^2}{16}+\frac{y^2}{b^2}$$=1$ and the hyperbola $\large\frac{x^2}{144}-\frac{y^2}{81} =\frac{1}{25}$ coincide. Then the value of $b^2$ is

$\begin{array}{1 1}(A)\;1 \\(B)\;5 \\(C)\;7 \\ (D)\;9 \end{array} $

1 Answer

Then the value of $b^2$ is 7
Hence C is the correct answer.
answered Apr 11, 2014 by meena.p
 

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