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  • Need solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Very short Answers Question 15 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Very short Answers Question 15 textbook solution.

Answers (1)

Answer : Slope of normal = 9

Hint :

\begin{aligned} &\text { slope of normal }=\frac{-1}{\text { slope of } \text { tan gent }} \\ &\text { i.e., } \frac{-1}{\frac{d y}{d x}} \end{aligned}

Given :

Given curve,

y=\frac{1}{x}

We have to write the slope of normal to the given curve at point \left(3, \frac{1}{3}\right)

Solution :

Here,

            y=\frac{1}{x}

On differentiating with respect to x, we get

   \frac{d y}{d x}=\frac{-1}{x^{2}}

Now,

\begin{aligned} &\text { slope of } \text { tan gent }=\left(\frac{d y}{d x}\right)_{(3, \frac{1}{3})}=\frac{-1}{9} \\ &\text { slope of normal }=\frac{-1}{\text { slope of tan gent }} \end{aligned}

                                     =\frac{-1}{\left(\frac{-1}{9}\right)}=9

Hence, the Slope of normal = 9

 

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