If the foci of the ellipse frac{x^2}{25}+frac | vedclass

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(B) The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Here $a^2=25$ and $b^2=16$. For an ellipse,$b^2=a^2(1-e^2)$,so $16=25(1-e^2)$,wh

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$5$

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(B) The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Here $a^2=25$ and $b^2=16$. For an ellipse,$b^2=a^2(1-e^2)$,so $16=25(1-e^2)$,which gives $e^2=1-\frac{16}{25}=\frac{9}{25}$,so $e=\frac{3}{5}$. The foci of the ellipse are $(\pm ae, 0) = (\pm 5 	imes \frac{3}{5}, 0) = (\pm 3, 0)$. The equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{b^2}=1$. Here $a^2=4$,so $a=2$. Since the foci of the hyperbola coincide with the foci of the ellipse,the foci of the hyperbola are $(\pm 3, 0)$. For a hyperbola,the foci are $(\pm ae_1, 0)$,so $ae_1=3$. Since $a=2$,we have $2e_1=3$,which means $e_1=\frac{3}{2}$. For a hyperbola,$b^2=a^2(e_1^2-1)$. Substituting the values,$b^2=4((\frac{3}{2})^2-1) = 4(\frac{9}{4}-1) = 4(\frac{5}{4}) = 5$. Thus,$b^2=5$.

If the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and the hyperbola $\frac{x^2}{4}-\frac{y^2}{b^2}=1$ coincide,then $b^2$ is equal to

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