Since, the curve $y=b \cdot e^{-x / a}$ intersects the Y-axis i.e., $x=0$. $\therefore \mathrm{y}=\mathrm{b} \cdot \mathrm{e}^{-0 / \mathrm{a}}=\mathrm{b} \quad\left[\because \mathrm{e}^0=1\right]$
Since, So, the point of intersection of the curve with Y-axis is $(0, b)$.

Now, slope of the given line $\frac{x}{a}+\frac{y}{b}=1$ at $(0, b)$ is given by $\frac{1}{\mathrm{a}} \cdot 1+\frac{1}{\mathrm{~b}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=0 \quad \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{b}}{\mathrm{a}}=\mathrm{m}_1 \quad$ [say $]$ Also, the slope of the curve at $(0, b)$ is given as: $\frac{d y}{d x}=\frac{-b}{a} e^{-x / a}$
$$
\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0, \mathrm{~b})}=\frac{-\mathrm{b}}{\mathrm{a}} \mathrm{e}^{-0}=\frac{-\mathrm{b}}{\mathrm{a}}=\mathrm{m}_2 \quad \text { [say] }
$$
Therefore, $\mathrm{m}_1=\mathrm{m}_2=\frac{-\mathrm{b}}{\mathrm{a}}$
Hence, the line touches the curve at the point, where the
curve intersects the axis of Y.