Show that the line $\frac{x}{a} + \frac{y}{b} = 1$, touches the curve $y = b \cdot {e^{ - x/a}}$ at the point, where the curve intersects the axis of $Y$.

We have the equation of line given by $\frac{x}{a} + \frac{y}{b} = 1$, which touches the curve $y = b \cdot {e^{ - x/a}}$ at the point, where the curve intersects the axis of $Y$ i.e., $x = 0$.

Therefore,$y = b \cdot {e^{ - 0/a}} = b$

So, the point of intersection of the curve with Y-axis is $(0,b)$.

Now, slope of the given line at $(0,b)$ is given by
$\frac{1}{a} \cdot 1 + \frac{1}{b} \cdot \frac{{dy}}{{dx}} = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{ - 1}}{a} \cdot b$

$\Rightarrow$ $\frac{{dy}}{{dx}} = - \frac{1}{a} \cdot b = \frac{{ - b}}{a} = {m_1}$ [say]

Also, the slope of the curve at $(0,b)$ is

$\frac{{dy}}{{dx}} = b \cdot {e^{ - x/a}} \cdot \frac{{ - 1}}{a}$

$\frac{{dy}}{{dx}} = \frac{{ - b}}{a}{e^{ - x/a}}$

${\left( {\frac{{dy}}{{dx}}} \right)_{(0,b)}} = \frac{{ - b}}{a}{e^{ - 0}} = \frac{{ - b}}{a} = {m_2}$

Since, ${m_1} = {m_2} = \frac{{ - b}}{a}$

Hence, the line touches the curve at the point, where the curve intersects the Y axis