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

$\therefore$ $y_1=b{e}^{\frac{{-x}_1}{a}}$ ......(i)

Also, the slope of the tangent to the curve is ${\left(\frac{dy}{dx}\right)}_{\left(x_1, y_1\right)}=\frac{-b}{a} e^{\frac{-x_1}{a}} =\frac{-y_1}{a}$ (by (i))

Now, the equation of tangent of the given curve at point $\left(x_1, y_1\right)$ is $y-y_1=\frac{-y_1}{a} \left(x-x_1\right)$

$\Rightarrow$ $\frac{x}{a}+\frac{y}{y_1}=\frac{x_1}{a}+1$

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 $(0, b)$.