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(A) The given curve is $y = b e^{-x / a}$.To find the point where the curve crosses the $Y$ axis,we set $x = 0$.Substituting $x = 0$ into the equation

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$x/a + y/b = 1$

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(A) The given curve is $y = b e^{-x / a}$.To find the point where the curve crosses the $Y$ axis,we set $x = 0$.Substituting $x = 0$ into the equation,we get $y = b e^0 = b$.So,the point of contact is $(0, b)$.Now,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$\frac{dy}{dx} = b \cdot e^{-x / a} \cdot (-1 / a) = -\frac{b}{a} e^{-x / a}$.At the point $(0, b)$,the slope $m$ is:$m = \left. \frac{dy}{dx} \right|_{(0, b)} = -\frac{b}{a} e^0 = -\frac{b}{a}$.The equation of the tangent line passing through $(x_1, y_1) = (0, b)$ with slope $m = -\frac{b}{a}$ is given by $y - y_1 = m(x - x_1)$.$y - b = -\frac{b}{a}(x - 0)$.$y - b = -\frac{b}{a}x$.Dividing both sides by $b$,we get $\frac{y}{b} - 1 = -\frac{x}{a}$.Rearranging the terms,we get $\frac{x}{a} + \frac{y}{b} = 1$.

The equation of the tangent to the curve $y = b e^{-x / a}$ at the point where it crosses the $Y$ axis is

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