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

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

Answers (1)

Answer:

                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

 

 

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