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

Let the point be $\left(x_1,y_1\right)$

$\therefore y_1=b{e}^{-x_1/a}$

Also, curve $y=b{e}^{-x/a}$

$\Rightarrow \frac{dy}{dx}=\frac{-b}{a}{e}^{-x/a}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{-b}{a}{e}^{-x_1/a}=\frac{-y_1}{a}$

Now, the equation of tangent of given curve at point 

$\begin{array}{r}\left(x_1,y_1\right)\text{ is }y-y_1=\frac{-y_1}{a}\left(x-x_1\right)\\ \Rightarrow \frac{x}{a}+\frac{y}{y_1}=\frac{x_1}{a}+1\end{array}$

Comparing with $\frac{x}{a}+\frac{y}{b}=1$

we get, $y_1=b$ and $1+\frac{x_1}{a}=1$

$\Rightarrow x_1=0$

Hence, the point is$(0, b).$