The point on the curve $\mathrm{y}={\mathrm{be}}^{\frac{-\mathrm{x}}{\mathrm{a}}}$at which the tangent drawn is $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$ is 

$$\begin{gathered} y=b{e}^{\frac{-x}{a}} \\ \frac{dy}{dx}=b{e}^{\frac{-x}{a}}\left(\frac{-1}{a}\right) \\ \frac{dy}{dx}=\frac{-b}{a}{e}^{\frac{-x}{a}} \end{gathered}$$

By Verification $\mathrm{m}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(0,\mathrm{b})}=\frac{-\mathrm{b}}{\mathrm{a}}$ and Slope of tangent is also $\frac{-\mathrm{b}}{\mathrm{a}}$ so,Option (1) Correct