#### Need solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Very short Answers Question 14 textbook solution.

Answer : Required angle $=\frac{\pi }{2}$

Hint :

1. If  $m_{1}=m_{2}$, they are parallel.
2. If  $m_{1} m_{2}=-1$, they are perpendicular to each other.

Given :

Given that the curve,

$y=e^{-x} \text { and } y=e^{x}$

We have to find the angle between the given curves at the point of intersection.

Solution :

Given,

$y=e^{-x}$                                                                                 ....(i)

$y=e^{x}$                                                                                    .....(ii)

From (i) and (ii), we get

\begin{aligned} & e^{x}=e^{-x} \\ \Rightarrow & x=0 \end{aligned}

Substituting the value of x in equation (ii), we get

$y=1$

So, the point of intersection of the two curves is $(0,1)$

On differentiating (i) with respect to x, we get

\begin{aligned} &\frac{d y}{d x}=e^{x} \\ &m_{2}=\left(\frac{d y}{d x}\right)_{(0,1)}=1 \end{aligned}

\begin{aligned} & \therefore & & m_{1} m_{2} &=-1 \\ \text { Since } & & m_{1} m_{2} &=-1 \end{aligned},

Hence, they are perpendicular to each other.

Hence, the required angle $=\frac{\pi }{2}$