Question

If the area of the region bounded by the lines $y-x=0$, $x=0$ and the normal to $y=e^x$ at $(0,1)$ is $\frac{1}{k}$, then k is equal to ________.

Answer 1

$y=x$ , $x=0$ and moved to $y=e^x$ at $(0,1)$

Let normal be $y=mx+c$

$\frac{dy}{dx}=e^x$ $\to$ tangent slope

normal slope $\to$ $=\frac{-dx}{dy}=-e^{-x}$

$m=-e^{-x}$

$m=-1$

$y=-x+c$ at $(0,1)$

$c=1$

Hence

Normal $y=-x+1$

Intersection between $y=x$ and $y=-x+1$

$=|\frac{1}{2},\frac{1}{2}|$

Hence area $=\frac{1}{2}\times 1\times \frac{1}{2}$

$=\frac{1}{4}$

$k=4$