If the foci of the ellipse frac{x^2}{16} + fr | vedclass

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(D) The given hyperbola is $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$.Dividing by $\frac{1}{25}$,we get $\frac{x^2}{144/25} - \frac{y^2}{81/25}

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$7$

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(D) The given hyperbola is $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$.Dividing by $\frac{1}{25}$,we get $\frac{x^2}{144/25} - \frac{y^2}{81/25} = 1$.Here,$a^2 = \frac{144}{25}$ and $b^2 = \frac{81}{25}$.The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{81}{144}} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$.The foci of the hyperbola are $(\pm ae, 0) = (\pm \frac{12}{5} 	imes \frac{5}{4}, 0) = (\pm 3, 0)$.For the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$,so $a = 4$.The foci of the ellipse are $(\pm ae', 0) = (\pm 4e', 0)$.Since the foci coincide,$4e' = 3$,so $e' = \frac{3}{4}$.For the ellipse,$e'^2 = 1 - \frac{b^2}{a^2}$.Substituting the values,$(\frac{3}{4})^2 = 1 - \frac{b^2}{16}$.$\frac{9}{16} = 1 - \frac{b^2}{16} \Rightarrow \frac{b^2}{16} = 1 - \frac{9}{16} = \frac{7}{16}$.Therefore,$b^2 = 7$.

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ are the same as the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,then $b^2 = \dots$

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