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

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(D) The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \frac{y}{b} = 0$. The product of the perpendicular distances from any point $(x_1, y_1)$ on the hyperbola to these asymptotes is given by the formula $\frac{a^2 b^2}{a^2 + b^2}$. Given that this product is $\frac{36}{13}$,we have $\frac{a^2 b^2}{a^2 + b^2} = \frac{36}{13}$. The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2} = \frac{a^2 + b^2}{a^2}$. Given $e = \frac{\sqrt{13}}{3}$,so $e^2 = \frac{13}{9}$. Thus,$\frac{a^2 + b^2}{a^2} = \frac{13}{9}$,which implies $9(a^2 + b^2) = 13a^2$,or $9b^2 = 4a^2$. From this,$b^2 = \frac{4}{9}a^2$,so $b = \frac{2}{3}a$. Substituting $b^2 = \frac{4}{9}a^2$ into the product equation: $\frac{a^2 (\frac{4}{9}a^2)}{a^2 + \frac{4}{9}a^2} = \frac{36}{13}$. $\frac{\frac{4}{9}a^4}{\frac{13}{9}a^2} = \frac{36}{13} \implies \frac{4}{13}a^2 = \frac{36}{13} \implies a^2 = 9 \implies a = 3$. Then $b = \frac{2}{3}(3) = 2$. Therefore,$a - b = 3 - 2 = 1$.

If the product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ to its asymptotes is $\frac{36}{13}$ and its eccentricity is $\frac{\sqrt{13}}{3}$,then $a - b =$

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