If the perpendicular distance from the focus of an ellipse $\frac{x^2}{9} + \frac{y^2}{b^2} = 1$ ($b<3$) to its corresponding directrix is $\frac{4}{\sqrt{5}}$, then the slope of the tangent to this ellipse drawn at $(\frac{3}{\sqrt{2}}, \frac{b}{\sqrt{2}})$ is
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The distance from a focus to the corresponding directrix in an ellipse is $a/e - ae$. The equation of the tangent at $(x_1, y_1)$ is found by the replacement $x^2 \to xx_1$, $y^2 \to yy_1$.