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(B) The given hyperbola is $16x^2 - 25y^2 = 400$,which can be written as $\frac{x^2}{25} - \frac{y^2}{16} = 1$. Here,$a^2 = 25$ and $b^2 = 16$,so $a =

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$\frac{320}{41}$

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(B) The given hyperbola is $16x^2 - 25y^2 = 400$,which can be written as $\frac{x^2}{25} - \frac{y^2}{16} = 1$. Here,$a^2 = 25$ and $b^2 = 16$,so $a = 5$ and $b = 4$. The equations of the asymptotes are $y = \pm \frac{b}{a}x$,i.e.,$4x - 5y = 0$ and $4x + 5y = 0$. Let $P(x_1, y_1)$ be any point on the hyperbola,so $16x_1^2 - 25y_1^2 = 400$. The lengths of the perpendiculars from $P$ to the asymptotes are $d_1 = \frac{|4x_1 - 5y_1|}{\sqrt{4^2 + 5^2}} = \frac{|4x_1 - 5y_1|}{\sqrt{41}}$ and $d_2 = \frac{|4x_1 + 5y_1|}{\sqrt{4^2 + 5^2}} = \frac{|4x_1 + 5y_1|}{\sqrt{41}}$. The product $p = d_1 d_2 = \frac{|16x_1^2 - 25y_1^2|}{41} = \frac{400}{41}$. The angle $	heta$ between the asymptotes is given by $	an \frac{	heta}{2} = \frac{b}{a} = \frac{4}{5}$. Therefore,$p 	an \frac{	heta}{2} = \frac{400}{41} 	imes \frac{4}{5} = \frac{80 	imes 4}{41} = \frac{320}{41}$.

If the product of the lengths of the perpendiculars from any point on the hyperbola $16x^2 - 25y^2 = 400$ to its asymptotes is $p$ and the angle between the two asymptotes is $\theta$,then $p \tan \frac{\theta}{2} =$

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