At which point does the line $\frac{x}{a}+\frac{y}{b}=1$ touch the curve $y=b{e}^{-x/a}$?

The line $\frac{x}{a}+\frac{y}{b}=1$ can be rewritten as $y=-\frac{b}{a}x+b$.The slope of this line is $m=-\frac{b}{a}$.For the line to touch the curve, the derivative of the curve must equal the slope of the line at the point of tangency.Curve: $y=b{e}^{-x/a}$ Derivative: $\frac{dy}{dx}=b{e}^{-x/a}\cdot \left(-\frac{1}{a}\right)=-\frac{b}{a}{e}^{-x/a}$ Setting the derivative equal to the slope:
$$-\frac{b}{a}{e}^{-x/a}=-\frac{b}{a}$$
$$e^{-x/a}=1⟹-\frac{x}{a}=0⟹x=0$$
Substituting $x=0$ into the curve equation:
$$y=b{e}^{-0/a}=b{e}^0=b$$
The point of tangency is $(0,b)$.