The value of b^2 such that the foci of the hy | vedclass

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(C) The given hyperbola is $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,which can be written as $\frac{x^2}{(12/5)^2} - \frac{y^2}{(9/5)^2} = 1$.

Vedclass · Accepted Answer

$7$

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(C) The given hyperbola is $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,which can be written as $\frac{x^2}{(12/5)^2} - \frac{y^2}{(9/5)^2} = 1$.Here,$a_1^2 = \frac{144}{25}$ and $b_1^2 = \frac{81}{25}$.The eccentricity $e_1 = \sqrt{1 + \frac{b_1^2}{a_1^2}} = \sqrt{1 + \frac{81}{144}} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$.The foci of the hyperbola are $(\pm a_1 e_1, 0) = (\pm \frac{12}{5} 	imes \frac{5}{4}, 0) = (\pm 3, 0)$.For the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$,the foci are $(\pm a_2 e_2, 0) = (\pm 4 e_2, 0)$.Since the foci coincide,$4 e_2 = 3$,so $e_2 = \frac{3}{4}$.For an ellipse,$e_2^2 = 1 - \frac{b^2}{a_2^2} = 1 - \frac{b^2}{16}$.Substituting $e_2 = \frac{3}{4}$,we get $\frac{9}{16} = 1 - \frac{b^2}{16}$.Thus,$\frac{b^2}{16} = 1 - \frac{9}{16} = \frac{7}{16}$,which implies $b^2 = 7$.

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

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