Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question Question 31 maths textbook solution.

Explain solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question, Question 31 maths textbook solution.

Answers (1)

Answer : (b) is the correct option

Hint : Multiply the slope of both curve

Given : x^{3}-3 x y^{2}+2=0 \text { And } 3 x^{2} y-y^{3}=2

Solution :

Given x^{3}-3 x y^{2}+2=0                                             (1)

Differentiating (1) we get

\begin{aligned} &3 x^{2}-3 y^{2}-3 x(2 y) \frac{d y}{d x}=0 \\ &3\left(x^{2}-y^{2}\right)=6 x y m_{1} \\ &m_{1}=\frac{x^{2}-y^{2}}{2 x y} \end{aligned}

And  3 x^{2} y-y^{3}-2=0                                                (2)

Differentiating (2), we get

\begin{aligned} &6 x y-3 x^{2} \frac{d y}{d x}-3 y^{2} \frac{d y}{d x}=0 \\ &6 x y=6 m_{2}\left(y^{2}-x^{2}\right) \\ &m_{2}=\frac{2 x y}{y^{2}-x^{2}} \end{aligned}

m_{1} \times m_{2}=\frac{x^{2}-y}{2 x y} \times \frac{2 x y}{y^{2}-x^{2}}=-1

Equation (1) and (2) at right angle

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