Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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

 

Posted by

infoexpert23

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question