If the product of eccentricities of the ellip | vedclass

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(D) For the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$,the eccentricity $e_1 = \sqrt{1 - \frac{b^2}{16}}$ (assuming $b^2 < 16$).For the hyperbola $\fr

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$\frac{144}{25}$

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(D) For the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$,the eccentricity $e_1 = \sqrt{1 - \frac{b^2}{16}}$ (assuming $b^2 For the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=-1$,we rewrite it as $\frac{y^2}{16}-\frac{x^2}{9}=1$. Here $a^2 = 16$ and $b^2 = 9$,so $e_2 = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.Given $e_1 \cdot e_2 = 1$,we have $\sqrt{1 - \frac{b^2}{16}} \cdot \frac{5}{4} = 1$.Squaring both sides: $(1 - \frac{b^2}{16}) \cdot \frac{25}{16} = 1$.$1 - \frac{b^2}{16} = \frac{16}{25}$.$\frac{b^2}{16} = 1 - \frac{16}{25} = \frac{9}{25}$.$b^2 = \frac{9 	imes 16}{25} = \frac{144}{25}$.

If the product of eccentricities of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=-1$ is $1$,then $b^2=$

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