Show that the line $\frac{x}{a} + \frac{y}{b} = 1$ touches the curve $y = b \cdot e^{-x / a}$ at the point where the curve intersects the $y$-axis.

(A) The curve is given by $y = b \cdot e^{-x / a}$.
To find the point where the curve intersects the $y$-axis,we set $x = 0$.
Substituting $x = 0$ into the curve equation,we get $y = b \cdot e^{0} = b \cdot 1 = b$.
Thus,the point of intersection is $(0, b)$.
Now,we find the slope of the tangent to the curve at $(0, b)$ by differentiating $y$ with respect to $x$:
$\frac{dy}{dx} = b \cdot e^{-x / a} \cdot \left(-\frac{1}{a}\right) = -\frac{b}{a} e^{-x / a}$.
At the point $(0, b)$,the slope of the tangent is $\left(\frac{dy}{dx}\right)_{(0, b)} = -\frac{b}{a} e^{0} = -\frac{b}{a}$.
Next,we find the slope of the given line $\frac{x}{a} + \frac{y}{b} = 1$.
Rewriting the line equation in slope-intercept form $y = mx + c$,we have $\frac{y}{b} = -\frac{x}{a} + 1$,which implies $y = -\frac{b}{a}x + b$.
The slope of this line is $m = -\frac{b}{a}$.
Since the slope of the tangent to the curve at $(0, b)$ is equal to the slope of the line,and the point $(0, b)$ lies on both the curve and the line,the line touches the curve at the point where it intersects the $y$-axis.