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

Toolbox:
• If $y=f(x),$ then $\bigg(\large\frac{dy}{dx} \bigg)_p$ = slope of the tangent to $y=f(x)$ at point $p$
Step 1
Given equation of the line is $\large\frac{x}{a}+\large\frac{y}{b}=1\: \: \: (1)$
Equation of the curve is $y=be^{-\large\frac{x}{a}}$
Since the curve intersects the $y$ axis the coordinate of $x =0$
$\Rightarrow y=be^{\circ}$
$=b$
$\therefore$ The points are $(o, b)$
Step 2
Consider curve $c_1$
$\large\frac{x}{a}+\large\frac{y}{b}=1$
differentiating w.r.t $x$ we get,
$\large\frac{1}{a}+\large\frac{1}{b}\large\frac{dy}{dx}=0$
$\Rightarrow \large\frac{dy}{dx}=-\large\frac{b}{a}$
Slope of the curve $c_1$ at $(o, b)$
$\bigg( \large\frac{dy}{dx}\bigg)_{(0,b)}=-\large\frac{b}{a}$
Consider the curve $c_2$
$y=be^{-\large\frac{x}{a}}$
differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=-\large\frac{b}{a}e^{-\large\frac{x}{a}}$
The slope of the curve at $(o,b)$ is
$\bigg( \large\frac{dy}{dx}\bigg)_{(o,b)}=-\large\frac{b}{a}e^{\circ}$
$= -\large\frac{b}{a}$
Hence it is clear that the $\bigg( \large\frac{dy}{dx}\bigg)_{c_1}=\bigg( \large\frac{dy}{dx}\bigg)_{c_2}$
Hence the curves touch each other at the point $(o,b)$