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(D) The given curve is $y = b e^{-x/a}$.Since the curve crosses the $y$-axis,we set $x = 0$. Substituting $x = 0$ into the equation,we get $y = b e^0

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$\frac{x}{a} + \frac{y}{b} = 1$

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(D) The given curve is $y = b e^{-x/a}$.Since the curve crosses the $y$-axis,we set $x = 0$. Substituting $x = 0$ into the equation,we get $y = b e^0 = b$. Thus,the point of contact is $(0, b)$.Now,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$\frac{dy}{dx} = b \cdot e^{-x/a} \cdot (-\frac{1}{a}) = -\frac{b}{a} e^{-x/a}$.At the point $(0, b)$,the slope $m$ is:$m = (\frac{dy}{dx})_{(0, b)} = -\frac{b}{a} e^0 = -\frac{b}{a}$.The equation of the tangent line at $(x_1, y_1)$ with slope $m$ is given by $y - y_1 = m(x - x_1)$.Substituting $(x_1, y_1) = (0, b)$ and $m = -\frac{b}{a}$:$y - b = -\frac{b}{a}(x - 0)$$y - b = -\frac{b}{a}x$$\frac{b}{a}x + y = b$Dividing both sides by $b$,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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