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

Answers (1)


                Slope of normal = -6


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


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 )


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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