Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 7 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 7 Maths Textbook Solution.

Answers (1)

ANSWER: The equation of normal y-a m^{3}=\left(\frac{-2}{3 m}\right)\left(x-a m^{2}\right)

HINTS:

Differentiating  with respect to  x to get its slope of tangent  2ay\left ( \frac{dy}{dx} \right )=3x^{2}

GIVEN:

ay^{2}=x^{3}at\; the \: point\left ( am^{2},am^{2} \right )

SOLUTION:

Upon differentiation

\frac{dy}{dx}=\frac{3x^{2}}{2ay}

m(tangent) at \left ( am^{2},am^{3} \right )is \frac{3m}{2}

The normal is perpendicular to tangent, therefore , m_{1},m_{2}=-1

m(tangent) at at \left ( am^{2},am^{3} \right )\: is \frac{-2}{3m}

The equation of Normal  is given by  ,

\Rightarrow y-a m^{3}=\left(-\frac{2}{3 m}\right)\left(x-a m^{2}\right)

Posted by

infoexpert21

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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question