$y = be^{-x/a}$
$\therefore \frac{dy}{dx} = -\frac{b}{a}e^{-x/a}$
Since the line $\frac{x}{a}+\frac{y}{b} = 1$ touches $\left(1\right)$
$\therefore -\frac{1/a}{1/b} = -\frac{b}{a} e^{-x/a}$
i. e., $- \frac{b}{a}= -\frac{b}{a} e^{-x/a}$
$\Rightarrow 1= e^{-x/a}$
$\Rightarrow - \frac{x}{a} = 0$
$\Rightarrow x = 0$
$\therefore y = be^{0} = 0 = b$
Hence reqd. point is $\left(0, b\right)$