#### Need solution for RD Sharma Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 10

Slope of normal $= -6$

Hint:

Slope of normal  $= \frac{-1}{\frac{dy}{dx}}$

Given:

Given equation of curve,

$y^{3}-x y-8=0$

To find:

We have to find the slope of normal to the given curve at the point $\left ( 0,2 \right )$

Solution:

Given curve,

$y^{3}-x y-8=0$                                                                                                                                 … (i)

On differentiating with respect to $x$, we get

$\Rightarrow 3 y^{2} \frac{d y}{d x}-x \frac{d y}{d x}-y=0$                                                                                                 $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

\begin{aligned} \Rightarrow & 3 y^{2} \frac{d y}{d x}-x \frac{d y}{d x}-y=0 \\ \Rightarrow &\left(3 y^{2}-x\right) \frac{d y}{d x}=y \\ \Rightarrow & \frac{d y}{d x}=\frac{y}{3 y^{2}-x} \end{aligned}

$\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(0.2)}=\frac{2}{3(2)^{2}-0}=\frac{1}{6}$

Hence the slope of the normal is $\frac{-1}{\frac{dy}{dx}}=-6$