Question
Write the angle between the curves $y = e^{-x}$ and $y = e^x$ at their point of intersections.

Answer

The given equation of curve are,
$y = e^{-x}...(1)$
$y = e^x...(2)$
Solving $(1)$ and $(2)$
$\text{y}=\frac{1}{\text{e}^\text{x}}=\frac{1}{\text{y}}$
$\Rightarrow\text{y}^2=1$
$\Rightarrow\text{y}=\pm1$
From $(2)$
$\pm1=\text{e}^\text{x}$
$\Rightarrow\text{x}=0$
So, the point of intersection is $P = (0, 1)$ and $Q = (0, -1).$
Slope of $(2)$
$\text{m}_2=\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}$
$\therefore\text{m}_2\text{ at }\text{P}=1\text{ and }\text{m}_2\text{ at }\text{Q}=1$
$\therefore\text{m}_1\times\text{m}_2=-1\times1=-1$
The angle between the curves is $90^\circ $

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