#### Need solution for RD Sharm Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 20

$y=0$  is the required equation of normal

Hint:

Equation of normal,

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

Given:

Given curve,

$y^{2}=8x$

To find:

We have to find the equation of normal to the given curve at the origin.

Solution:

We have,

$y^{2}=8x$

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

$\Rightarrow 2 y \frac{d y}{d x}=8$                                                                                                                         $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{8}{2 y} \\ &\Rightarrow \frac{d y}{d x}=\frac{4}{y} \\ &\Rightarrow-\left(\frac{d x}{d y}\right)=\frac{-y}{4} \end{aligned}

At origin i.e. $\left ( 0,0 \right )$

Here the equation of normal,

\begin{aligned} &\left(y-y_{1}\right)=-\left(\frac{d x}{d y}\right)_{(0.0)}\left(x-x_{1}\right) \\ \Rightarrow \quad &(y-0)=0(x-0) \\ \Rightarrow \quad & y=0 \end{aligned}

Hence the required equation of normal is $y=0$.