Question

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

Answer 1

Given curve $y=b{e}^{\frac{-x}{a}}$

Differentiate w.r.t $x$

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

When the curve crosses Y-axis, $x=0$

At $x=0\Rightarrow y=b$

So the curve cuts Y-axis at $(0,b)$

At $(0,b)$ , $\frac{dy}{dx}=\frac{-b}{a}{e}^0=\frac{-b}{a}$

Now equation of tangent to $y$ at $(0,b)$ is

$\Rightarrow y-b=\frac{-b}{a}(x-0)$

$\Rightarrow y-b=\frac{-bx}{a}$

$\Rightarrow bx+ay=ab$

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

Hence proved.