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(C) For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the asymptotes are given by $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \frac{y}{b}

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$\frac{36}{13}$

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(C) For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the asymptotes are given by $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \frac{y}{b} = 0$.Here,$a^2 = 9$ and $b^2 = 4$,so $a = 3$ and $b = 2$.The asymptotes are $\frac{x}{3} - \frac{y}{2} = 0$ or $2x - 3y = 0$ and $2x + 3y = 0$.Let $P(x_0, y_0)$ be any point on the hyperbola,so $\frac{x_0^2}{9} - \frac{y_0^2}{4} = 1$.The perpendicular distance $d_1$ from $P$ to $2x - 3y = 0$ is $d_1 = \frac{|2x_0 - 3y_0|}{\sqrt{2^2 + (-3)^2}} = \frac{|2x_0 - 3y_0|}{\sqrt{13}}$.The perpendicular distance $d_2$ from $P$ to $2x + 3y = 0$ is $d_2 = \frac{|2x_0 + 3y_0|}{\sqrt{2^2 + 3^2}} = \frac{|2x_0 + 3y_0|}{\sqrt{13}}$.The product $d_1 d_2 = \frac{|(2x_0 - 3y_0)(2x_0 + 3y_0)|}{13} = \frac{|4x_0^2 - 9y_0^2|}{13}$.Since $\frac{x_0^2}{9} - \frac{y_0^2}{4} = 1$,we have $4x_0^2 - 9y_0^2 = 36$.Thus,$d_1 d_2 = \frac{36}{13}$.

The product of the perpendicular distances drawn from any point on the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ to its asymptotes is

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