#### Need solution for RD Sharma Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 17

Equation of normal,  $y=3x-5$

Hint:

Use equation of normal,

$\left(y-y_{1}\right)=-\frac{d x}{d y}\left(x-x_{1}\right)$

Given:

Given curve,

$x^{2}+y^{2}-2 x+4 y-5=0$

To find:

We have to find the slope of normal to the given curve at point  $\left ( 2,1 \right )$

Solution:

Given equation is

$x^{2}+y^{2}-2 x+4 y-5=0$

On differentiating both side with respect to $x$ , we get

$\Rightarrow \quad 2 x+2 y \frac{d y}{d x}-2+4 \frac{d y}{d x}=0$                                                                                         $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

$\Rightarrow \quad(y+2) \frac{d y}{d x}=1-x$     [Taking common $2$  from each term]

\begin{aligned} &\Rightarrow \quad \frac{d y}{d x}=\frac{1-x}{y+2} \\\\ &\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(2,1)}=\frac{1-2}{1+2}=\frac{-1}{3} \\ \end{aligned}

$\Rightarrow \quad-\left(\frac{d y}{d x}\right)_{(2,1)}=\frac{1}{3}$

$\Rightarrow \quad-\left(\frac{d x}{d y}\right)_{(2,1)}=3$

Now, equation of normal is

$\left(y-y_{1}\right)=-\frac{d x}{d y}\left(x-x_{1}\right)$

Substituting these value, we get

i.e.   $y_{1}=1, x_{1}=2, \frac{-d x}{d y}=3$

\begin{aligned} &(y-1)=3(x-2) \\ \Rightarrow & \quad y-1=3 x-6 \\ \Rightarrow & \quad y=3 x-5 \end{aligned}

Hence, $y=3 x-5$ is required equation of normal.