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(D) Given the curve $y = be^{-x/a}$.To find the point where it crosses the $y$-axis,set $x = 0$.Then $y = be^0 = b$.So,the point of intersection is $(

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

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(D) Given the curve $y = be^{-x/a}$.To find the point where it crosses the $y$-axis,set $x = 0$.Then $y = be^0 = b$.So,the point of intersection is $(0, b)$.Now,find the derivative $\frac{dy}{dx} = b \cdot (-\frac{1}{a}) e^{-x/a} = -\frac{b}{a} e^{-x/a}$.The slope of the tangent at $(0, b)$ is $m = \left(\frac{dy}{dx}\right)_{(0, b)} = -\frac{b}{a} e^0 = -\frac{b}{a}$.The equation of the tangent line at $(x_1, y_1) = (0, b)$ with slope $m = -\frac{b}{a}$ is given by $y - y_1 = m(x - x_1)$.Substituting the values: $y - b = -\frac{b}{a}(x - 0)$.$y - b = -\frac{b}{a}x$.Multiply by $a$: $ay - ab = -bx$.Rearranging: $bx + ay = ab$.Dividing by $ab$: $\frac{bx}{ab} + \frac{ay}{ab} = \frac{ab}{ab}$.Thus,$\frac{x}{a} + \frac{y}{b} = 1$.

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

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