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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.

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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