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(B) Given,the hyperbola is $16 x^2 - 9 y^2 = 1$. This can be written as $\frac{x^2}{(1/4)^2} - \frac{y^2}{(1/3)^2} = 1$. Here,$a^2 = \frac{1}{16}$ and

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$4 e_2$

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(B) Given,the hyperbola is $16 x^2 - 9 y^2 = 1$. This can be written as $\frac{x^2}{(1/4)^2} - \frac{y^2}{(1/3)^2} = 1$. Here,$a^2 = \frac{1}{16}$ and $b^2 = \frac{1}{9}$. The eccentricity $e_1$ of the hyperbola is given by $e_1^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{1/9}{1/16} = 1 + \frac{16}{9} = \frac{25}{9}$. Thus,$e_1 = \frac{5}{3}$. The eccentricity $e_2$ of the conjugate hyperbola is given by $e_2^2 = 1 + \frac{a^2}{b^2} = 1 + \frac{1/16}{1/9} = 1 + \frac{9}{16} = \frac{25}{16}$. Thus,$e_2 = \frac{5}{4}$. Now,$3 e_1 = 3 	imes \frac{5}{3} = 5$. Also,$4 e_2 = 4 	imes \frac{5}{4} = 5$. Therefore,$3 e_1 = 4 e_2$.

If $e_1$ and $e_2$ are the eccentricities of the hyperbola $16 x^2 - 9 y^2 = 1$ and its conjugate respectively,then $3 e_1 = $

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