#### Please Solve RD Sharma Class 12 Chapter15 Tangents and Normals Exercise Fill in the blanks Question 9 Maths Textbook Solution.

Point  $=\left(\frac{-1}{2}, 0\right)$

Hint:

1. Using equation of tangent is $\left(y-y_{1}\right)=\frac{d y}{d x}\left(x-x_{1}\right)$
2. Tangent line meets the  $x-$axis. i.e. $y= 0$

Given:

Given equation of curve,

$y=e^{2 x}$

To find:

We have to find the point where curve at  $\left ( 0,1 \right )$ cut $x-$axis.

Solution:

The given equation of curve,

$y=e^{2 x}$                                                                                                                                                … (i)

Differentiating equation (i), we get

\begin{aligned} &\Rightarrow \quad \frac{d y}{d x}=2 \cdot e^{2 x} \\\\ &\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(0.1)}=2 \cdot e^{2(0)}=2 \end{aligned}

Slope of tangent to the curve =2

So equation of tangent is

$(y-1)=2(x-0)$

$\Rightarrow \quad y=2 x+1$

Above tangent line cuts the $x-$  axis, where  $y= 0$

$\therefore \quad x=\frac{-1}{2}$

Hence the required point is  $\left ( \frac{-1}{2} ,0\right )$